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Coordinates of Relative Maxima or Minima

Date: 31 Dec 1994 17:58:13 -0500
From: Anonymous
Subject: math problem test

Hi ..... this is just a test.....
 I heard you solve math problems... so I'm just playing......

given  f(x) = 15 x^(2/3) + 5x, find coords of relative max or min. 
and for what x values is the function concave down?  



Date: 1 Jan 1995 20:31:51 -0500
From: Dr. Ken
Subject: Re: math problem test

Hello there!

Yes, we're a service to students in kindergarten through 12th grade.
Students send along math problems that puzzle them, and we try to 
help them out.  Spread the word!

As for your question, here's how you would proceed.  You'd take the 
first derivative of f(x), and then look at which values of x make f'(x) 
zero, and for which values of x the derivative doesn't exist.  These will 
be the candidates for local (relative) maxima and minima.

We have f(x) = 15x^(2/3) + 5x.  So we get f'(x) = 10/[x^(1/3)] + 5.  
If you're not sure how I got that, let us know.  So then you set this
expression equal to zero, and then solve for x.  In this case, we get 
x=-8. And the derivative doesn't exist when x=0.  So these are our only 
candidates for local maxima or minima.  Plugging in to f, we see that 
the coordinates of these points are (-8, 20) and (0,0).

But we still need to test whether these really ARE local extrema.  So 
let's do it.  The best way is to take the second derivative of the original
function, plug in our x-value, and see whether a positive or a negative
number pops out.  If it's positive there, the function is concave up 
(think of an upward wind - the positive direction - blowing on the 
graph of the function), and if it's negative, it's concave down (a 
downward wind).

Since you also want to know all values of x for which this function 
is concave down, we'll kill two math birds with one stone.

The second derivative of your function is f"(x) = -10/[3x^(4/3)].  So 
when is this less than zero?  We'll set up an inequality to find out.  

-10/[3x^(4/3)] < 0
      3x^(4/3) > 0
       x^(4/3) > 0

Great.  This works out nicely.  Sometimes it gets really messy when 
you have fractional exponents.  Anyway, when you raise something to 
the 4/3 power, you take the cube root of it (which you can do with any 
number, positive or negative) and then you raise the result to the fourth 
power.  So you'll end up with a positive number all the time.  Which 
means this condition on the second derivative is fulfilled all the time, 
i.e. that the function is concave down everywhere.

Except zero.  See, that's the tricky part of this problem.  Notice that we
can't plug in zero to our second derivative (or even the first derivative,
for that matter), since we'd then divide by zero.  So we'll have to check
zero by hand.  It looks like zero is a local minimum, since f(0)=0, and 
if you plug in anything just to the left or right of zero you get something
positive, i.e. bigger than zero (do you know how you could check that 
and make absolutely sure?).  So zero is a local minimum.

So here's what we've got.  We've got a function that's concave down
eveywhere except one point, and it has a local maximum at (-8, 20) 
and a local minimum at (0, 0).  What a neat problem.  Note the little 
sharp point at (0, 0); that's why the derivative doesn't exist there.

Anyway, if you have more questions, please feel free to write back!

-Ken "Dr." Math
Associated Topics:
High School Calculus

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