# End Point Maxima and Minima?

• Oct 23rd 2011, 01:37 PM
Justinram22
End Point Maxima and Minima?
Hello!

Quick question, in my Calc class we are covering End Point Maxima and Minima.

I understand that for lets say prove there is a max at x=-2

f(x)=(x+2)^(3/2)+5*x^2 for x >= -2

f'(x)=(3/2)*(x+2)^(1/2)+10*x

You can just plug in -2 and get -20, which proves that it is a max because the function is decreasing to the right of it.

What I do not understand is this

f(x)=(25-x^2)^(1/2)-2*x^2 for -5 <= x <= 5

f'(x)=(1/2)*(2x)*(25-x^2)^(-1/2)-4x

It is asking if 5 is a max or a min

But when you plug in 5 it DNE

The book describes using infinity, but I do not understand how they are using it?

Thanks for anyone who can offer any help!
• Oct 23rd 2011, 02:18 PM
skeeter
Re: End Point Maxima and Minima?
$f(x) = \sqrt{25-x^2} - 2x^2$

$f'(x) = \frac{-x}{\sqrt{25-x^2}} - 4x$

note that $f(x)$ is defined at $x = \pm 5$ , but $f'(x)$ is not.

$f'(x) = 0$ at $x = 0$

for $-5 < x < 0$ , $f'(x) > 0$ ... $f(x)$ is increasing on that interval. therefore, $f(-5)$ is an endpoint minimum.

for $0 < x < 5$ , $f'(x) < 0$ ... $f(x)$ is decreasing on that interval. therefore, $f(5)$ is an endpoint minimum.

$f(0)$ is a maximum