Hello guys! I have h[x]= cube root of x and I need to find absolute extrema. Please help

Question

mathteacher1729 · Answer

Over what interval?

anonymous · Answer

great thanks, from -1 to 8

anonymous · Answer

I know that I start with taking the derivative for the first derivative test

anonymous · Answer

h'[x] = 1/3x^-2/3 right?

anonymous · Answer

but from there i'm kinda in the dark.

mathteacher1729 · Answer

Find all points where the first derivative is equal to zero. Those are critical points.

anonymous · Answer

thats what I dont understand I don't know how solve 1/3x^-2/3 for 0  :(

mathteacher1729 · Answer

$$\huge f(x) = x^{1/3}\qquad f'(x) = \frac{1}{3}x^{-2/3}$$
When does $f'(x) = 0$ ?
$$\huge 0 =\frac{1}{3}x^{-2/3}$$ multiply by 3 on both sides to get 
$$\huge 0 = x ^{-2/3} \Rightarrow  0 = \frac{1}{x^{2/3}}$$
That is never true (can't have 0 in the denominator). So there are no critical points. 
 That means you must test the endpoints next.