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!

Re: End Point Maxima and Minima?

note that is defined at , but is not.

at

for , ... is increasing on that interval. therefore, is an endpoint minimum.

for , ... is decreasing on that interval. therefore, is an endpoint minimum.

is a maximum