So, last class we have taken a numerical example to illustrate the solution of optimal control problem using the calculus of variation. And you see this is the simple circuit is there our initial condition of the circuit is given like this way voltage across the capacitor is given 2 volt and current flowing to the inductor is 0 compare. So, I have to find out the control law u of t means current flowing to the main circuit in such a way. So, that at time t is equal to t f the current voltage across the capacitor will be 0 and current flowing to the inductor will be 0 find out the corresponding control law u of t. In other words I of t such that the loss in the circuit is minimum that was to the least possible energy power loss in the circuit is minimum. So, this thing we have converted this description of this circuit is converted into a dynamic equation and x 1 dot is equal to x 1 dot is equal to x 2 and x 2 dot is equal to x 1 minus x 1 plus u of t. This is the state phase depth description and it is a general structure is x dot is equal to f function of x t u t d t which is in this particular case it has become into a state phase form x dot is equal to a x of t b u of t. So, we know the our dynamic equation of this one our problem is to minimize this performance index that one. So, this we are solving this problem by using it is a constant optimization problem we have converted into a what is called unconstraining optimization problem. And using the what is called in state of Lagrangian function we are using the what is the Hamiltonian function that necessary condition for the Hamiltonian function is once you form the Hamiltonian function that necessary condition for the Hamiltonian function is first is del h del u is equal to 0. Then you have a del h del lambda is equal to x dot then next is del h del x dot is equal to minus lambda dot that three necessary equation necessary condition we have to solve it in addition to the what is called the boundary conditions. And our boundary condition if you see we have shown you earlier that our boundary condition that h del h that is what we have written see this one this is the boundary condition. And we have to use this boundary condition corresponding to our problems since that what is called del t f because we know at what time this that final state of the circuit that means voltage across the capacitor and current flowing the inductor will be 0 that t f is fixed. So, and now so we know the final state of this one is we know x voltage across the capacitor is 0 and current flowing to the inductor is 0. So, you know the final state of the system that means delta x f is equal to 0 that means delta but t f is not 0 you find out the t f for which the system final state which reached to the origin mean 0 0 position. So, we have seen by using the three necessary conditions as well as the boundary condition we have come to this point that what is called lambda 2 0 cos t f lambda 1 0 sin t f these are the three unknowns are there, but this equation 18 contains the three unknowns lambda 1 of 0 lambda 2 of 0 and lambda sin t f t f which is not known, but our main aim if you see our problem is to find out the u control law by solving that equation. So, u 2 you need the description of lambda 2 of t lambda 2 of t. So, now first lambda 2 of t if you see the expression for lambda 2 of t you got it lambda 2 of t expression that this is the expression lambda 2 of t expression. So, here unknowns are lambda 2 0 lambda 1 0. So, this we must know this one. So, we know this equation and another two equation because three unknowns are there another two equation we get from the equation of the state equation trajectory that x 1 t expression and x 2 t expression. The 15 corresponding to x 1 t and 17 corresponding x 2 of t if you put the t is equal to t f we will get the expression for that one what we are going to write it here. From 15 t f t is equal to t f from 15 and 17 at t is equal to t f from 15 and 17 at t is equal to t f we will get x 1 t f 15 is the expression for x 1 t expression. We are putting t is equal to t f in that expression 15 expression.  So, lambda 0 of t t f then sin t f see the equation 15 just now I told you that equation 15 you just see this is the equation 15 last class we have seen equation in that expression we are putting t is equal to t f t wherever t is that t is equal to t f final time this one. So, we got it is similarly from equation 17 this one I will put t is equal to t f t is equal to t f in the left hand side and right hand side both expression this.  We want to find out a control law e u in such a way that resistive loss in the circuit is minimum. And not only that current flowing at a time t is equal to t f that current will be 0 through the inductor is 0 that means current flowing through the inductor is 0. We want to find out a control law e u in such a way that resistive loss in the circuit is minimum and not only that current flowing at a time t is equal to t f that current will be 0 through the inductor and voltage across the inductor capacitor will be 0. So, these things now we have a let us call this equation is 19 and this equation is 20. So, now see this one there are 3 equations 18 to 20 3 equation 3 unknowns, but unfortunately this equation are the non-linear algebraic equation. So, you have to solve it by numerical techniques you can solve it. So, you are writing solving 18 to 20 one can get one can get the solution for lambda 1 of 0 lambda 2 of 0 and t f. Once you know that lambda 1 of 0 and lambda 2 of 0 then our problem is solved because we can find out our control law.  And lambda 2 of t is nothing but a lambda 2 of t you see the expression for lambda 2 of t here lambda 2 of t is lambda 2 of 0 sin t lambda 1 of 0 cos t sin t. So, this values are now known. So, lambda 2 of t we know intern u of t we know once you know u of t and lambda u of t then we can get the response for optimal trajectory for x 1 t x 2 of t see x 1 of t is what we got it. So, our x 1 of t this. So, I know lambda 1 of 0 lambda 2 of 0. So, I know x 1 of t trajectory of that one similarly x 2 of t putting these values I know the response of this one. So, these are the solution for this we can get it. So, or one can use to solve the equation 18 to 19, 18 to 20, 18, 19, 20 that is a set of non-linear algebraic equation one can solve by using a software generally used in MATLAB software what is called MATLAB toolbox you can use it to solve this set of equation. So, by using MATLAB toolbox that is a one solver is there f solver agree you can use this one to get these 3 values 3 unknown values lambda 1 of 0 lambda 2 of 0 and t f we can solve this one. So, now what is the optimal control law just now we have seen our optimal control law is equal to optimal control law u of t is nothing but a minus lambda 2 of t c from equation 13. So, lambda 2 expression we know. So, lambda 2 expression we know that lambda 2 0 cos t minus lambda 1 0 sin t. So, that this is the lambda 2 expression see this expression. So, first is this is equation number 4 this is equation number 4 and this is expression lambda 2 expression from 13 see the equation 13 this is the lambda 2 expression. So, once you know this one this is known all are known by solving this set 3 set of equations. So, and we have to find out from t 0 to t f we know at what time t f we know at what time t is equal to f this final state of the system. That means, voltage across the capacitor will be 0 and voltage across voltage not voltage current flowing to the inductor will be 0 at time t is equal to t f if you solve this 3 set of non-linear algebraic equations. So, if you see the block diagram of this one what we got it this we have a see this one our system x dot x 2 dot of t we have integrated the output of the integrated is x 2 of t and this is initial condition of this x 2 of 0 is equal to 2. Then this x 2 do is nothing but x 1 dot of t x 1 dot of t is nothing but x 2 integrated to realize that one is x 1 of t.  So, minus x 1 of t so there is a another is b this is our b and this is our u star and where from u star is getting this is u of t and this is b is in your case 0 1 this and what is this this is our optimal control law optimal control law. And how what is this expression u of t is equal to just now we have seen it u of t is that one. So, what is this one lambda 1 of 0 sin t I am writing from this one u of t sin t minus lambda 2 0 cos t. So, you see this control law is the open loop control law. So, it entirely depends on the solution of lambda 1 0 and lambda 2 0. So, this so optimal so it depends on the initial value of say u u depends on initial value of u of t. So, this is the open loop it does not depends on the state information to control the systems. So, it is a open loop what is called system and the control law is generated by with this expression. And we have seen how lambda 1 0 and lambda 2 0 are obtained from this solution of this one. So, lambda 2 0 if you see this one then when we are solving this equation. So, this initial condition lambda 2 how you are solving this equation lambda 2 expression is this one alpha 1 plus alpha 2 we know this lambda 2 of 0. So, whole control law depends on the initial condition of that alpha 1 0 and alpha 2 0. So, it is a open loop solution of this one and this is dependent on the initial condition. And in practice it is not acceptable that control law. So, what is this our sufficient condition and sufficient whether the system that objective function is minimized or maximized along this trajectory or whatever the control law we have generated you start whether the corresponding performing index minimized or maximized that has to be checked with this sufficient condition. So, this is equal to x square of t minus del square h del x of t del u of t similar del square of h del x of t del u of t then del square h this del x of t del x of t del x u square of t.  And if it is greater than 0 for then you conclude implies that j minimum we got it if it is less than 0 for j maximum that you have to check that things. So, you know how to solve this problems of that one next we will come to the solution of what is called linear quadratic regulator problem with finite time. So, next is before that we will discuss the what is the performing index says performance indices. Performing index is a measure of system performance is a measure of system performance first suppose a our objective. So, it is a measure of system performance first suppose that an objective function that our objective objective is to if our objective is to control dynamic response of the system of the system that described by x dot is equal to A x of t plus B u of t and initial condition is known or given let us call equation number one and our output equation u of t is equal to C x of t this is the output equation. So, this is the equation what is called state equation and this is the this is our output equation so, that output equation on a fixed interval fixed interval t 0 to t f. So, that one what is our objective the component of the state variables the component of the state variables x are small our objective is to get the component of the components of the state variables are small. So, for that one so, our problem is you design a control law u such that our component of the state variables are small for that one what should be the choice of our performing index. Since, component of the state variables are small the performing index what we have selected in quadratic form that means, sum of the error square must be small. So, a suitable suitable performance index a suitable performance index performance index is selected as so, our performing index j 1 is equal to t 0 to t f because our over the time interval t 0 to t f that our component of the state variable should be small. So, that is can be represented by x transpose t into x of t d t should be small that this indicates over the time interval t 0 to t f that this and nothing but a what is called a clean and norm or distance of the vector that is nothing but a each component if it is x 1 x 2 dot x n the dimension is x n is nothing but a x 1 square plus x 2 square plus x 3 square dot dot x n square this represents. So, this should be a small so, you have to find out the control law in such a way so that j is small means each component of the state variable is small. So, that for that one what should be our control law. So, if you want to design a control law there is no restriction on the input. So, input magnitude can be very high which is not acceptable in practical situation. So, this is one of the our performing index we consider if our each component of the state variables if you want to make it small that corresponding performing index is that one. Now, this or one can write it this also one can write it by giving weightage on the state suppose our q is the diagonal matrix it indicates if you expand this one if q is the diagonal matrix then it is nothing but a x 1 square multiplied by small q 1 and diagonal elements of q if it is a small q 1 q 2 dot dot q n. Then if you expand this one you will get x 1 square into q 1 x 2 square into q 2 that means we are giving weightage on this performing index with this one that is also could be a one of the one of the performing index when our objective to make the each component of the state variable is each component state variable to make small. This is the either one of this will be our choice second performing index if our motivation is to each component of the control variable that means you want to make it what is called large not too large if you want to make it each component of the each component of the input variable not to make too large then what should be the choice of our performing index. So, the component of the input variables u of t and u of t dimension is m cross 1. So, we have a m components that each component when to make it not to make very large and not too large then our suitable choice of performing then a suitable choice of performing index performance index performance index is j 2 is equal to t 0 to t f over the interval and 0 to t d t t transpose this d t d t.   So, u 2 t u transpose u indicates the energy or control afford. So, what is the energy is delivered this indicates physical interpretation is what is the energy delivered to the system to achieve our objectives our objective is now in this case to make the input variables are each input variable not too large again that is we have to make it or one can select this one t 0 to t f u transpose giving some weightage on u. So, this u of t d t this is also may be our choice of performing index. So, our problem is like this way find out the control law u in such a way this performing index is minimized subject to the constraint the x dot is equal to a x plus b u, but here we do not have any control on the state. State magnitude may be a very large all these things again. So, here we do not have any control on the state. So, now question is so this is the let us call this is 3 this is 4. So, now we make it comment that what is the minimization of 4 means the minimal energy to the system because this is because this is because input energy is related to the input generality related to the because this is because because input is generally related to the energy consumed in the circuit or in the system.  So, it is nothing but a control effort or control energy involved in the system again control energy is supplied to the system this one this indicates. So, if you see first case when the system is in equilibrium position what we expect we expect our state should be in equilibrium position in linear system the generally our equilibrium position is 0 0. So, if you give a perturbation with the initial condition the state will deviate from the equilibrium position and the corresponding deviation you see our x minus this is the our state x minus x 0 is the our equilibrium position this transpose into x minus x t minus 0 this is the our equilibrium position from there we have given some perturbation. So, this interpreted x transpose of x. So, this is to minimize this one, but you see whenever minimizing this one whenever minimizing that quantity then we do not have any control on input whenever minimizing that second part then we do not have any control on the state. So, there is a what is called so remark it may be pointed out we cannot simultaneously minimize the performance index indices 1 that is we are consider 1 that is j 1 of dot and 2 j 2 of dot. That means this is nothing but a t 0 to t f x transpose of t x of t d t and this is nothing but a t 0 to t f u transpose t u of t d t we cannot simultaneously minimize because why you cannot do it. Because the minimization of j of j 1 results large control effort or control signal control effort also you can control signal while minimization of while minimization of j 2 results minimization j 2 results small control signal. So, this is the minimization of j 1 because we do not have any control. So, if you give it physically if you give a pumping more energy to the system that quantity will be minimized. Second part of this one is when we are minimizing that one when we are minimizing the results small control signal minimization of this one results small control signal. So, these two things simultaneously we cannot minimize because it is a contradict of one another. So, what you have to do this one we have to make some to this dilemma we have to make some compromise. So, the compromise we can make it by using the convex combination of these two performance index. So, to solve this dilemma we could compromise between two conflicting objective functions objectives by minimizing by minimizing the performance index performance index that is performing index that is a combination of that is a combination of that is a convex combination of convex combination of j 1 and j 2. So, our resultant performance index is j dot is lambda into j 1 dot and plus 1 minus j 2 1 minus lambda is j 2 and lambda varies from 1 to n 1 1 to n when lambda is equal to 1 then this is minimization of this one when lambda is equal to 0 then minimization of j 2. So, you are taking the convex combination of these two things. So, one can go one can get good results good response of the system by making trial and error giving more weight is here or less weight is here depending upon the situation with lambda value we can get the tune the system response. So, let us now see how to solve our control problem, but what do you mean by the linear quadratic regulator problems on finite time linear quadratic regulated problem and regulator you mean optimal control. So, linear quadratic regulator problem what we understand. So, describe the system equation like this later that the system is described as x dot is equal to A x of t plus B u of t. So, that is this one and y is equal to y of t is equal to c x of t and this is the initial state x t of 0 is equal to x t of 0 either initial state is given to you this is equation number 1 this is equation number 2 and this dimension is n cross 1 this input is m cross 1 and output is p cross 1 p outputs are there. So, our problem is to find a control law the statement of the problem or problem is to find a control law u star of t such that the associated associated performance index.  Such the associated performance is minimized over the interval t 0 to t f. So, our problem is to find a control law u such that this the performance index is minimized and our state is moved the initial state the control law will drive the initial state from x of t 0 to from initial state x t 0 x t 0 to drive the final state x is equal to t f near to the what is called origin or near to the 0. So, that is our problem. So, once again I state the problem like this way that our problem is find out the control law u of t in such a way. So, this performance index is minimized. In other words you can say you find a control law u that will drive the state from the initial state x to the final state x t f which is close to the origin or 0 for that you find out the control law u of t. And what is the performance index is there similar to this one is a terminal cost similar to earlier discussion it is a terminal cost and this whole part is a integral cost functional this is the integral cost functional.   And q is the waiting matrix associated with the state vector and this r also symmetric matrix r is equal to r transpose and it is a only positive definite matrix not positive semi definite matrix. So, this indicates that this performance index you find out the control law that will drive find out the optimal control law that will drive the state from initial state this to a final state x t f which is near to the origin by minimizing this performance index. And each term of the performance index has a physical significance because our original system is at rest whatever the position is there and we have given the system perturbation x t is equal to t 0 something. So, that state will deviate from the equilibrium position and this indicate the sum of the deviations squares with some weight is q.  So, simultaneously we want to minimize both in such a way that we find you minimize that u you minimize that performance index j such that that control f out u will drive the state from x t 0 to x t f which is near to the origin. So, this is called finite time regulator problem within a finite time the state should reach to the origin near origin.  Now, only difference you should there is a half in present in this expression this half is intentionally we kept it we know if the objective function if you multiply it by a scalar quantity the optimal point does not change it. But, optimal value of the function value will change it. So, just I multiply it by half the optimal value at which the objective function will be maximized on minimize that value will not change optimal point at which the function value is minimum maximum will not change. So, why I made it half you will see when you will do the that find out the necessary condition of this objective function there are two terms a two will come from the when you will differentiate with respect to x all with respect to y all these things with respect to lambda we will do just like a your what is called necessary condition what you got it the del h del u the two term two will come into the picture and this two term and half will cancel. If you do not keep it half the two will be carried out throughout the derivations that is only this. So, intentionally we make it half in the expressions. So, let us why we have introduced half we just write it here remember the half in the cost function is associated mainly to cancel to that would have otherwise otherwise been carried on throughout the that result or derivations. So, if you just see our main problem of this our the statement problem our problem our aim this is the important thing our aim at driving the initial state x of t 0 is equal to x 0 to the smallest possible value near to 0 in the interval t 0 to t f, but without but without spending too much control effort too much control effort. Means, you would transpose you would transpose you to achieve the goal to achieve the goal. So, this is our statement of the problem if you see this one our problem is find out the control law u of t such that our original state will with the application of this control law our original state will drive to a what is called at time t is equal to t f x t x t f which is near to the origin will drive not only this it will satisfy our original system dynamic equations. So, this is our so now this problem is solution of this problem is easy the same as what we have described earlier now our the state equation we have now described in state in state of what is called that dynamic equation is expressed in in terms of state equations. This is only difference is this one now that physical significance of this weighting matrix this is called state weighting matrix q is called the state weighting matrix. The q is whose dimension is n cross n q is a q transpose agree and which is a positive semi definite matrix and it is called state weighting matrices. Then r is equal to r transpose which is greater than 0 when positive definite positive this is positive semi definite is called control weighting matrices agree. So, if you want to keep the state small in the expression if you see in the expression if you want to keep the state small then choice of q must be positive definite matrix. So, our implication you can say implication of various weighting matrices and what is this f of t t f is the this is positive semi definite matrix and this is the terminal weighting matrix terminal cost weighting matrix.  So, it is n cross n so next is implication of various weighting matrices. So, first implication q of t state weighting matrix this is a state weighting matrix agree to keep the state small agree because initially the state small the system state at the equilibrium position if you give a perturbation to the system the state will deviate from the equilibrium position. So, if you want to keep the state small then the choice of then the integral of the expression half x transpose q x of t this should be x transpose q x of t this should be non negative non negative small implies that q must be greater than 0. You want to keep the this integral of this one small this should be a non negative small quantity the q must be a positive definite matrix. So, how you can make it see this expression how can you make it this q this part is small by giving more control effort to the system the u magnitude should be that u star u t of u t multiplied u t transpose of u t should be large then it will drive the state to small value of this one. But, when you will give the control effort large then what will happen this control effort is not permissible to act into the system because it is a physical system. So, you cannot make it too large of this one second implication of that is R R with R R of t the control weighting matrix to keep to keep the control effort to keep the control effort small the integral part integral of the expression half x transpose u transpose this R u of t should be should be positive small why positive small this quantity that that what with this quantity is you are getting this quantity. If it is 0 then u transpose u t if it is a 0 that there is we are not applying to any control effort to the system. So, it will not be able to control the state of this one. So, this should be a positive definite. So, this input positive implies that R should be greater than 0. So, another important issue is most of the practical control problems you will find there is a restriction on the control input. That means the control input has a minimum value and maximum value of this one. So, this is most of the practical problem we have a restriction on the control input, but for the time being we will discuss that there is no restriction on the control input. The control input u of t there are no constraint imposed on the control input u of t in your statement of the problem. We have not mentioned any control constraint in the statement of the problem which is very uncommon in real practice or which is very important in practice to design the control systems to design the close loop optimal configurations. First we will design a linear optimal control linear regulator problem without constraints. Then we will take that if there is a constraint in the control input constraint the input as well as in more practical problem constraint in the input as well as constraint in the state. So, that problem we will discuss later. So, the last one is what is called infinite time infinite time infinite final time that means t f tends to infinity. When the final time t tends to infinity the terminal cost tends to infinity the terminal cost when final time t tends to infinity that terminal cost does not have any sense because at time t is equal to infinity x of t f will approach to 0 or it will go to asymptotically it will go to 0. So, in that situation the terminal waiting function has no what is called sense at all. So, we will put that f f t f is equal to 0 when t tends to infinity. So, it is called this type of problem infinite final time this type of problem is called what is called infinite horizon problems infinite infinite horizon control problem. So, another two points are there we will discuss less next class of this one. So, infinite time regulator problem t tends to infinity you see the state will reach to the origin means near equilibrium position at that situation the terminal waiting function has no sense. So, we will assign that t f is equal to 0. So, that type of problem is called what is called infinite horizon control problems. So, we will


Welcome to the review for exam 4 for Math 1210 Calculus 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. Now, if you're watching this video, you've probably participated in the previous exams for this lecture, for this course. Therefore, you're probably familiar with the basic policies, rules, procedures, format of this exam, so I'm not going to go into all the details about that. If you do have questions about that, I would suggest consult the core syllabus, the announcements and policies on Canvas or just reach out to your instructor. But again, many of us are probably familiar with that. Exam 4 is going to be covering lectures 31 through 41 in our lecture series. So this will include the topics of extreme values, both absolute extrema and local extrema, the mean value theorem, the first and second derivatives test, L'Hopital's rule, the idea of curve sketching using derivatives to sketch a curve, solving optimization problems, approximating values using linearization and Newton's method, and an introductory topic on antiderivatives. That is, can we compute a few basic antiderivatives? This test will include 15 questions. The first 10 questions will be about the multiple choice questions, for which you'll get five points each for those. Questions 11, 12, 13, 14 and 15 will be in the free response section. Their points do vary on the low side. Question 11 is worth seven points on the high side. Question 15 is worth 13 points. Each question will be approximately 10 points in value, on average. And then also there will be two points for turning in your approved notes at the end of this exam. In terms of when or where you'll take this exam, and like I said, other semester-to-semester policies, rules about the exam, please can contact the course syllabus to find that information. In this video, I want to review the basics of the types of questions that you should be preparing for before you take the exam. This is based upon, of course, the practice exam that's available to students. You can find it also in the link to this page. All right, so if we go through the questions in the order they're presented, question number one will give you a question about critical numbers. So remember what a critical number is. A critical number is a number in the domain of the function that makes the derivative equal to zero, or it makes the derivative undefined. So what are those values? Again, what are those values that make the derivative undefined or the derivative equal to zero? So these are what we call critical numbers. Critical numbers are important for optimization. They're important about monetization and concavity. And so at the beginning of this unit, we introduced the idea of critical numbers. So if you want, if you want some practice on how to compute critical numbers or just the definitions, please consult the first lecture in this unit. Go back to lecture 31. Basically, it's going to have you calculate a derivative. So the derivative rules that you know about already, the power rule, the product rule, quotient rule, chain rule, all of those derivative rules from our third unit are very much applicable in this section, in this chapter, in this test, I should say. Feel free to use those. You have to do a derivative calculation, but then apply it to find these critical numbers. Question number two, you'll be asked to compute a limit, but this limit should be a fairly easy application of L'Hopital's rule. You'll notice, for example, with this limit, take the limit as x approaches zero, e to the x minus one over x. If you were just to plug in x equals zero, you'll get e to the zero minus one over zero, which e to the zero, of course, is equal to one. You get one minus one over zero. This looks like zero over zero. So this is a type of function, a limit for what I should say that L'Hopital's rule applies. So remember that when you have an indeterminate form of the form zero over zero or infinity over infinity, then L'Hopital's rule tells us that the limit as x approaches a, where a could be a finite number like zero, positive or negative, but could also be infinite plus or minus infinity. We see that the limit of f of x over g of x will be the same as the limit as x approaches a of f prime of x over g prime of x. So apply L'Hopital's rule to help you out with the calculation number two. Question number three, will be a question that involves your knowledge of the first derivative test. Before we go on, I should mention that L'Hopital's rule, at least the easier questions were introduced in section 34. About half of that lecture was about L'Hopital's rule. The other half was about the second derivative test. L'Hopital's rule was also covered in lecture 35 as well. So consult those of you want more practice with L'Hopital's rule. Returning to question number three, like I said, question number three will be some usage of the first derivative test. Which remember what the first derivative test is telling us. The first derivative test tells us that if we have some critical number, right, and we know that the derivative goes from a positive to a negative. Well, if the derivative is positive, that means the function's increasing. And if the derivative is negative, that means the function's decreasing. So we're seeing something like this. Oh, if the first derivative switches from positive to negative, that indicate we have a local maximum on the graph. On the other hand, let's say we have our critical number again. But we see some other effects. Let's say the derivative goes from negative to positive at that critical number. That means the function would switch from decreasing to increasing for which the graph would have to look something like this. Oh, this indicates we have a local minimum. This is what the first derivatives test is all about. Also, like again, like I said, when the derivative is negative, the function's decreasing. When the derivative is positive, the function's increasing. So this function right here is given to us as a polynomial function. Find the intervals for which f is decreasing. So what you would want to do is calculate the derivative of f and identify probably via a sign chart. Where is the derivative negative? If you're looking for the where the function was increasing, we would calculate the derivative figure out where it's positive, you might be asked to identify maxima or minima use the first derivative test to help you out on a question like number three, I'm going to erase some of this so we can see the next question. But before we go on, let me do remind you that the first derivative test was introduced at our lecture series in lecture 33. Please consult that lecture, the notes, the videos and the company homework if you need some more practice sections of the book for more practice on that. Question number four, we'll ask you to compute an anti derivative. So anti derivatives are the very last unit we talked about, excuse me, the last section we talked about in this unit. So anti derivatives were talked about in lecture 41. So we're looking for a function whose derivative is given. So do we can we find a function whose derivative is cosecant cotangent? Can we find a function whose derivative is two times e to the two theta where we also have some addition here. So look at the anti derivative rules that we've learned about in lecture 41. So there is a anti derivative version of the power rule, the sum rule, the difference rule, things like that. Every every derivative formula we know can be reversed into an anti derivative formula as well. These will be very, very basic. We will talk some more about anti derivatives in unit five of our course. And also in calculus two, you talk so much more about anti derivative. So this is just a very elementary, just a very basic anti derivative calculation. Don't forget the plus C right? Because again, we're looking for the most general anti derivative. Maybe you think the answer is the first one listed cosecant theta plus e to the two theta that that maybe that's good. Maybe this is this is an anti derivative of such a function. I'm not saying it is maybe it is maybe it isn't. But what I can tell you is that choice A has to be wrong because it's not the most general. The difference between say a and D. Excuse me. That's cotangent a and is it on there? I don't see it anywhere. So apparently a is wrong for other reasons. But if you were to compare, for example, F and C right here, those are the same except for that C is missing that plus C. So this is this is a anti derivative maybe of R, but it's not the most general. We want the family of anti derivative. So don't forget your plus C. I've seen many students who've written on a no card. Remember the plus C do whatever you need to to help you remember that. But that's going to be a big deal. On this one, it's multiple choice. You can see the plus C is hopefully that'll be enough of a Q. But you do need to know you do need to know that the plus C in order to get full credit on this one right here. So moving on to the last question on the first page here is question number five, which is going to be about Newton's method. Newton's method we talked about in lecture 40 in our series lecture 40 also talked about linearization, which we won't need that on this question. This question is going to be a question on Newton's method, for which basically what you need to know about Newton's method is the formula for Newton's method. So you need to know that you have this sequence of numbers that converges towards the value you're trying to approximate that the term X in plus one is equal to X in minus F of X in over F prime of X in. So in particular, if you're given X one and you're given the function F, you hopefully should be able to find X two because X two will equal X one minus the function evaluated at X one divided by the derivative evaluated X one. And typically for this type of question, like we saw in the lecture, like we saw in the homework, you really should use computer approximations, a computer algorithm to help you with this because these calculations can get very tedious very quickly. But for this test, you won't have access to computer software or graphing calculator. You'll have access to just a scientific calculator or less. And so I do like to ask questions about Newton's method, but it's going to just be one iteration. So if I give you X one and the function, can you calculate X two? If you can calculate X two from X one, then ideally you could calculate any value in the sequence. But again, to avoid tedious calculations, I don't need you to calculate X 10 X two we sufficient. So you do need to know Newton's formula. So put that on your formula sheet. Moving on to the next page. Question number six, this will be a baby optimization type question. So this one we're asked to find the minimum perimeter of a rectangle with an area of 200 centimeters squared. So the fact that you see that there's a minimum perimeter, that's an optimization type problem. No doubt about that. Just as a reminder, optimization questions we talked about in lectures 38 and 39 in our lecture series. So look at some of the examples you can see from those ones right there. We had many examples that had to do with geometry of some kind. So you should know either by memory or put them on your no card. You should know basic formulas of perimeter and area of two dimensional figures that will include rectangles, triangles, circles, other types of polygons. I'm not necessarily going to pull out the most exotic things at all, but you should definitely know things like the Pythagorean equation. Some of these basic two-dimensional formulas. You will also see an optimization question, the free response. This one again is on the easier side. Fairly straightforward, fairly routine, not very exotic compared to what you might see later on. But do be prepared, there will be two optimization questions on this test. One of the free response, one in the multiple choice section, multiple choice section, that's right. So question number seven, this is going to be a question about the second derivative test. We had a question earlier about the first derivative test. This one is the second derivative test. Now, I want to well, one, first remind you that second derivative test was introduced in lecture 34. So consult the lecture notes, the lecture videos, your own notes, homework, textbooks, for more examples for that. But question seven, much like we saw earlier on the previous page with question number three about the first derivative test, it is intended to use the second derivative test. But hey, wait a second, how am I going to know that necessarily is this a multiple choice question? Well, one way to know that is because if I ask you a question about concavity, that has to do with the second derivative. Remember that if your second derivative is positive, that means that your function is going to be concave upward. And if your second derivative is negative, that means your original function will be concave downward. And whenever you get a sign change of the second derivative, it switches from positive to negative or vice versa. If you ever see a sign change that implies you have some type of point of inflection. Alright, what makes the second derivative equal to zero or undefined gives us these potential points of inflection. So you might be asked questions like that. You also might be told things like the following. Let's say that our function, let's say let's look at a variable, let's say that f prime of say zero is equal to zero and f double prime of zero is equal to negative one. What can we say about the point? Well, since f prime of zero is zero, that means it's a critical number, so it could be an extremum. But then that it's negative, the function's concave down at that situation. If you're concave down at a critical number, oh that means you are a local maximum. So you could say that x equals zero is a maximum. Even if you don't know what the function is, maybe the function's not given to, you're just told that the first derivative is zero and the second derivative is negative one. But let me give you a variation of that. What if we tell you that the first derivative at zero is equal to two and that the second derivative at zero is equal to say negative two? Well, the function's increasing because the derivative's two. The function is concave downward because the second derivative is equal to negative two. Is it a maximum? The answer is nope. Is it a minimum? The answer is nope. The thing is the fact that it's not a critical number, we know that it can't be an extremum. So we could say something like that. But what if we did one other variation? What if the derivative was equal to zero? So it's a critical number. But then what if the second derivative was also equal to zero? Then it's like oh no decision can be made because the second derivative is inconclusive in that situation. So be aware that on this question number seven, also number three, with the first derivative test, you might not actually be given a function to calculate the derivative of or in this case the second derivative. You might just be given information on the derivatives and then you're supposed to infer information at extremum or concavity, monotonicity from that information. So you do need to know the first and second derivative test. They are slightly different and you will be asked questions requiring the use of one or the other. So you need to be prepared to do both. Question number eight is going to be a question about linearization. Remember the linearization is just a code word for the tangent line. That is the tangent line approximation for what you'll remember from our lecture series that the linearization l of x, well basically it's just the tangent line. You take y minus f of a is equal to f prime at a times x minus a. Then you solve for y and so you end up with f prime of a times x minus a plus f of a where this is the x coordinate of the point of tangency. This is the y coordinate of the point of tangency and this is the slope of the tangent line at the point of tangency. So the linearization. What we've seen is that if x is approximately a that implies that l of x will be approximately f of x where there's some epsilon delta stuff going on there but we don't have to worry about that. Just if x is close to a then we can infer that l of x will be close to f of x. It's great. So you have to use the linear approximation here. Now let me be warned or let me warn you I should say that if you're asked to approximate something like the square root of 65 I'm aware that you have a scientific calculator. Could you plug that into your calculator? Well your calculator is probably going to give you a decimal expansion because it's an irrational number. And so if you're like I'm going to work this thing backwards I'm going to take answer f maybe because when you plug that in that gives you the decimal expansion that's closest to what your calculator says the square root of 65. I'll tell you right now that's a trap and the fact that you're watching this video means you're trying to avoid such traps right. The thing is the linearization is an approximation technique but your scientific calculator is going to use things more sophisticated than the linearization technique. Probably something like Newton's method or something better than Newton's method is what your calculator is going to be using. And therefore the best answer that is the answer that's the closest to the square root of 65 is not what the linearization will give you and therefore you can get the wrong answer if you don't use the linearization technique. So you now have been warned. The answer is not about the approximation the answer is about the technique. And so using a different technique might give you again a more a better approximation but not the approximation linearization gives you. So again you've been warned here. Question number nine is going to be another question about Lopital's rule. This one is going to be on I'd say the harder side compared to what we saw in question number two. I've realized I forgot to mention that linearization showed up in lecture 40. I told you part of the lecture 40 was about linearization part most of it was about Newton's method. So go back to lecture 40 if you need some more practice on linearization. All right. Lopital's rule remember was explained in lectures 34 and 35. The harder ones of course were found in lecture 35 and so that's what you should expect right here in question number nine. Question number nine for example if you just plug in x equals pi you're going to end up with pi from the left minus pi and then you're going to get times by cosecant excuse me cosecant of pi from the left for which if you take pi minus pi that's going to give you zero. It is a little bit smaller than zero right so you get zero minus and then for the for cosecant you're going to get some type of vertical asymptote. Notice that pi makes cosine go to zero excuse me make sine go to zero because cosecant is one over sine. So it makes so cosine pi goes to z excuse me sine goes to zero at pi which makes cosecant undefined. Basically you're going to get some type of plus or minus infinity. We could actually figure out what exactly it is. Is it positive infinity or negative infinity because we're approaching pi from the left but I don't want to tell you that because actually it doesn't matter in this situation. This is since you get since you have this like zero times infinity form because whether it's plus or minus infinity the sine doesn't matter. This is something you need to use L'Hopital's rule for but this is not L'Hopital directly. L'Hopital directly applies to zero over zero infinity over infinity because this is zero times infinity you have to modify you have to switch from zero times infinity to one of these forms somehow and that's what was talked about in lecture 35. So look at those examples and those appropriate homework questions if you need some more practice on that. We now arrive to the last question the multiple choice section question number 10. This is a question which you will be given a graph of the derivative. This is a very important thing there's not going to be variation here. You will be given the graph of the derivative not the graph of f the graph of f prime. So when you see this function right here this is the graph of f prime but then you're asked where is f increasing and this is where you have to remember that f increases exactly where excuse me f is f is increasing exactly where the first derivative is positive. So you're looking not for where the function is increasing because excuse me you're not looking for where the derivative is increasing so this is a red herring right you're like oh the derivative is increasing here and it's increasing here and so you might be tempted to say something like negative infinity to zero union zero or two. Uh-uh wrong though this is not the right answer this is where the derivative is increasing where is the function increasing the function will be increasing when the derivative is positive and likewise the function is decreasing when the derivative is the first derivative excuse me is negative. I could also ask you where is f concave upward right? Uh if the function if the function's concave upward that means the second derivative is positive but if the second derivative is positive that means the function f is f prime excuse me the function f prime is increasing so this this right here this interval which we had earlier with c this is where f is concave upward this is not where f is increasing so be careful about that you'll be you will be given the derivative the graph of f prime what information about f can we infer from the graph of f prime now this is stuff we've been talking about through the entirety of this fourth unit in particular emphasis should be placed on lectures 33 about the first derivative test 34 about the second derivative test I would also include lectures 36 and 37 about curve sketching and also the very first section about 31 because again this whole unit is about how does information about the derivative affect the function and so in particular we saw questions like this from lectures 34 and 33 so let's move on to the free response section what type of questions are we going to see here so question number 11 it's worth seven points you'll be given a so-called extreme value problem extreme value problem these were problems this was like the predominant problem we saw in lecture 31 remember the extreme value theorem told us that if we have a continuous function on a closed interval which is exactly what we have right here then that function has a absolute maximum and an absolute minimum so you need to identify what the absolute extrema for this function what what these extrema are because it will be provided that is they will exist and you need to find it that question is worth seven points the second free response question on this page question number 12 it'll be worth eight points and on this question you're going to see a question about antiderivatives but in the context of science of some kind so notice in this question you're given the velocity function of in this case a fallen ball you're given the initial height of it and then you're asked to see how high it is layer on so you're at you're asked what is s of 10 right well the the idea to take care of here is that s of s of t for example is the integral of the antiderivative of v of t right so you have to calculate the antiderivative to find the general antiderivative then you use the initial value to find this particular antiderivative that applies to the situation right here so these were some of the story problems that we saw at the end of this unit in particular lecture 41 was about antiderivatives so be prepared to answer some type of story problem basically a scientific application a scientific interpretation of these antiderivatives so you're given you'll be given the functions derivative you need to figure out what the original function is and answer questions about that so we have to calculate the antiderivative velocity to get position a similar question is I could give you marginal cost so that you could say something about cost using the relationship of antiderivatives question number 13 will be a question about the mean value theorem okay so the mean value theorem was given to us really early on in our unit it was lecture 32 which the mean value theorem says that if we have some function right if it's sufficiently continuous and sufficiently uh differentiable I'll let you look up the exact assumptions necessary then the then if you have two points right so here's our point x equals a and x equals b then there's that you take the ck line between those that somewhere between there between a and b there will be a tangent line that is parallel to them so some interpretations of the mean value theorem so question number 13 is a conceptual type question which asks you to use the mean value theorem again in some content conceptual situation this function right here it gives you a specific function um you're supposed to explain why the hypotheses of the mean value theorem are satisfied on the interval zero to one and therefore the mean value theorem guarantees the existence of such a point you're supposed to find where is that point or explain why it doesn't exist if it doesn't apply uh some other variants we saw like thinking of the homework for lecture for homework 32 another question we could see something like oh prove that this equation has at most one solution that was another application of the mean value theorem so question number 13 will ask you to do something similar to the homework uh for lecture 32 homework 32 question number 14 will be that will be the second optimization question that I promised you earlier optimization just as a reminder occurred in lecture 38 and 39 this one will be on the harder side maybe like three-dimensional questions will be involved I do not expect everyone to have these various three-dimensional formulas memorized like this one has to do let's see what are we trying to do here we're trying to minimize the cost of some type of can some which is a cylinder basically that has a constraint on volume well you are provided the volume of the cylinder okay so you don't need to have that memorized so things like volume of a comb pyramid cylinder even a sphere I'm not going to expect you to know all those things the volume of like a rectangular prism length times width times height I don't think that's unreasonable you should know that one but you'll be provided these three dimensional formulas if they're there this one will be harder than what you saw on the multiple choice because again I intend for you to show your work so go through the steps of an optimization problem tell me what is the constraint what is the optimizing the optimizing function give me the derivative of the optimizing function plug in the constraint find the critical numbers there's a lot to be shown here this question's worth 10 points we now arrive upon the final question on this exam question for 15 it is worth the most 13 points this will be a curve sketching problem you'll be given a function the algebraic formula for that function and you'll be expected to graph it right here on these grid lines but there are a lot of steps you need to show to get full credit so you need to mention to me what is the domain of this function what makes this function defined what makes it undefined what are the intercepts both x or y intercepts check for symmetry so if you take f of negative x do you get negative f of x so it's an odd function do you get f of x so it's an even function or do you get something else so it's neither do check for symmetry there's points worth there mention any discontinuities if the function has any does the function have a remove point does it have a vertical asymptote does it have a jump discontinuity I want to see the in behavior of the function what's the limit as x approaches infinity what's the limit as x approaches negative infinity does it have a horizontal asymptote does it have an oblique asymptote maybe the function's domain is restricted like the natural log in which case you take the limit as x approaches infinity and as x approaches 0 from the right do so appropriately now for your sake and also my sake as the grader here you will be given the first derivative it's going to be computed and factored so hopefully since most of the legwork has been taken away from you now you can then use the first derivative to set up a sign chart like here are my critical numbers and it goes positive negative positive negative positive this is our first derivative so what does it say about the functions like up it's increasing it's decreasing it's decreasing maybe there's some discontinuity there maybe it's increasing what have you so fill out this sign chart and tell me the intervals where the function's increasing decreasing mentioned in the local extreme authority the same thing will also happen for the second derivative tests for concavity you will be given the computed factored second derivative so you don't actually have to compute derivatives on this question you have to interpret the derivatives that's given to you so again build a sign chart determine where the function concave up where the functions concave down identify any points of inflection and so these are all the parts the way you show your work then I want to see all of this stuff come together into this picture the domain needs to be what the picture the picture of the domain should be what you calculated all of the x intercepts should be represented here the symmetry discontinuity should be listed here the in behavior the concavity the monotonicity all of that stuff should be listed in this graph and that's why this question's worth so many points there are a lot of pieces going on here as a reminder the lectures about curve sketching were lectures 36 oh boy 36 and 37 so please look at the examples and homework for that to better prepare yourself for this test so that then concludes our review for exam number four we've talked about all the different question types clearly specific functions that you see right here could change very likely to change but the format of these questions will be very similar to what you saw in this review if you do have any other questions beyond what we've talked about in this video please reach out to me your instructor either by email come by face come by office hours ask me a question in class post a question on the discussion board or whatever other format is most comfortable to you I'm here to answer any questions you have so please please let me know if you have any questions and I will hopefully see you next time as we continue our study of calculus bye everyone