need help with absolute extrema, know the answer, need to know why.
f(x)=3x^2/3 +x,; [-9, 1]
f'(x)=2x^-1/2 +1

Question

need help with absolute extrema, know the answer, need to know why.
f(x)=3x^2/3 +x,; [-9, 1]
f'(x)=2x^-1/2 +1

anonymous · Answer

max is -8, min is 0

anonymous · Answer

take the derivative, find the critical points (zeros of the derivative, or where it is undefined) then check at those numbers and the endpoints of the interval

anonymous · Answer

I see 1 is also a max...

anonymous · Answer

i think your derivative might be wrong

anonymous · Answer

$$f(x)=3x^{\frac{2}{3}} +x$$
$$f'(x)=2x^{-\frac{1}{3}}+1$$

anonymous · Answer

$$f'(x)=\frac{2}{\sqrt[3]{x}}+1$$
undefined at x = 0 so that is one critical point

anonymous · Answer

That is what I have..., now I need solve the derivitive to equal 0, but that is undefined

anonymous · Answer

right

anonymous · Answer

$$\frac{2}{\sqrt[3]{x}}+1=0$$
$$2=-\sqrt[3]{x}$$
$$x=-8$$

anonymous · Answer

so two critical points are at x = 0 and x = -8
need to check those two along with the endpoints of the interval

anonymous · Answer

the endpoint 1 is also a max
so I plug the critical points into the function right?

anonymous · Answer

yes

anonymous · Answer

you need 
$$f(-9),f(-8),f(0),f(1)$$ smalles it min, largest is max

anonymous · Answer

*smallest

anonymous · Answer

you got it from here?

anonymous · Answer

Thanks
I think I got it, for now

anonymous · Answer

yw