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?
 = \sqrt{25-x^2} - 2x^2)
 = \frac{-x}{\sqrt{25-x^2}} - 4x)
note that )
is defined at 
, but )
is not.
 = 0)
at 
for 
,  > 0)
... is increasing on that interval. therefore, )
is an endpoint minimum.
for 
,  < 0)
... is decreasing on that interval. therefore, )
is an endpoint minimum.
)
is a maximum
