Question

Consider the function f(x)=x(√25−x^2),−1≤x≤5 This function has an absolute maximum value equal to?

Answer 1

first find the critical points:
f'(x)=sqrt(25)-x^2-2x^2=sqrt(25)-3x^2=5-3x^2

f'(x)=0
5-3x^2=0
3x^2=5
x^2=5/3
x=sqrt(5/3) or x=-sqrt(5/3)
notice that -sqrt(5/3) is out of the interval -1<=x<=5 so we'll disregard that....
now we plug in the values sqrt(5/3) and the endpoints of the interval which are -1 and 5 and whichever gives us the highest value of f(x) is the absolute maximum value
f(-1)=4 f(5)=-100 f(sqrt(5/3))=4.30331482911
so we can see that the maximum of f(x) on interval -1<=x<=5 is at x=sqrt(5/3)

Answer 2

But (√25−x^2) is all under one square root. That's my problem.

Answer 3

in that case
f'(x)=(25-2x^2)/(sqrt(25-x^2))

f'(x)=0
(25-2x^2)/sqrt(25-x^2)=0
25-2x^2=0
2x^2=25
x^2=25/2
x=5/sqrt(2) x=-5/sqrt(2)
-5/sqrt(2) is out of the interval [-1,5] so we'll disregard it..
plug in 5/sqrt(2), -1, and 5 and whichever gives the highest value is the maximum..
f(5/sqrt(2))=6.25 f(5)=0 f(-1)=-sqrt(24)
so it will have maximum at x=5/sqrt(2)