Application of first derivative: extrema of h(t) = t^2/3 - 4

[cmnalo]

I need to find the relative extrema, if any, of h(t) = t^2/3 - 4

h'(t) = 2/3t^-1/3

x = 0 is the only critical point (correct?)

h(0) = -4

How do I conclude that this is the max or min?

Any help would be appreciated.

Answer is: No RMax, RMin: h(0) = -4

 

[skeeter]

first derivative test ... check the sign of h'(t) on both sides of t = 0.

if h'(t) changes from (+) to (-), then h(0) is a max
if h'(t) changes from (-) to (+), then h(0) is a min

the first derivative test should be in your text ... look it up to understand why it works.