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Find derivatives and bounds of x^2 e^(-3x)

  1. Nov 16, 2007 #1
    1.with the function: f(x) = x^2e^-3x

    a)find f'(x) and f''(x)
    b)find the critical points of f(x)
    c)use the derivative to determine where f(x) is increasing and decreasing.
    d)use the second derivative to determine where f(x) is concave up and concave down. Identify any points of inflection.

    I got the first and second derivative, I just want to make sure I took the derivative correctly.

    f'(x)= 2x(e^-3x) - (3x^2)(e^-3x)
    f''(x)= 2(e^-3x) + (9x^2)(e^-3x)

    Then I set the first derivative equal to zero in order to find the critical points.

    getting x= 0 and x= 2/3

    I want to make sure I did a and b correctly, then I need some help with c and d.


    2. For the function y= axe^-bx^2 use calculus to find exact values of a and b so that the function has a maximum at (2,1).

    I tried taking the derivative and then plugging in the x and y values, but I'm having trouble finding a and b.

    y'= a(e^-bx^2) - 2bxax(e^-bx^2) then plugging in (2,1)
    1= a(e^-b(2)^2) - 2b(2)a(2)(e^-b(2)^2)

    I'm lost after that part.

    3. Use calculus to find the best bounds for f(t)= tsqrt(4-t^2) for -1 =< t =< 2 (-1 less than or equal to t, and t less than or equal to 2.

    I believe you take the derivative and then set it equal to zero to get the critical points. After that, I need some help to get the bounds.

  2. jcsd
  3. Nov 16, 2007 #2


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    1c) For a function to be increasing, the first derivative must always be positive(i.e. >0) and for a fn. to be decreasing it must be...guess it....that's right always negative...

    concave up and concave down simply means max or min points...so sub the critical values into f''(x) and if you get a +ve answer,then it is a min point and if -ve..a max point.
  4. Nov 16, 2007 #3
    ok thanks, I've got #1 done. I still need help with #2 and #3. If anybody can help, it would be appreciated. thanks
  5. Nov 16, 2007 #4


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    Well if it has a max at (2,1) this means that when x=2 [itex]\frac{dy}{dx}=0[/itex] and (2,1) is on the curve...so you can make 2 equations in a and b, so you can solve them and get the values
  6. Nov 16, 2007 #5
    so step 1. find derivative and equal it to zero? 2. I'm having trouble making the two equations.

    the derivative so far is y'= ae^-4b - 4b2ae^-4b= 0
    Last edited by a moderator: Nov 16, 2007
  7. Nov 16, 2007 #6


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    well you can just divide by e^(-4b) as this will never be equal to zero for any value of b

    well since (2,1) lies on the curve...sub those points into the eq'n and check the other equation

    ALSO: is it [itex]e^{-bx^2}[/itex] ?
  8. Nov 16, 2007 #7
    Well the original equation is y= axe^-bx^2

    so I took the derivative and then plugged in 2 into the x's, which gave me e^-4b, that's where I got the e^-4b, but i'm not sure if that's right.

    So, eliminating e^-4b in the equation I should have a - 4b2a?
  9. Nov 16, 2007 #8
    I would suspect that if everything is right so far (the derivative i mean) now its pretty easy just take in the values (2,1) plug them in the original equation and you get another equation in terms of a, b plus the one that you now have from the derivative you get a system of two equations and two unknowns a, b ... you can find the values of a, b.

  10. Nov 16, 2007 #9
    ok so the original equation with the values plugged in, is

    1= 2ae^-4b

    But what i'm still fuzzy on is, do I plug (2,1) into the derivative as well?
  11. Nov 19, 2007 #10
    This is your teacher. I know who you are. Which part of the "I did not receive help from anyone" did you not understand of the take home test?
  12. Nov 19, 2007 #11


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