Associated Topics || Dr. Math Home || Search Dr. Math

### 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?

thanks

Jill
```

```
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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search