find the critical points of h(t) = (t^2-4)^(1/3)

To find the critical points of the function h(t) = (t^2-4)^(1/3), we need to find the values of t where the derivative is either zero or undefined. Critical points occur at the values of t where the function may have local extrema (maximum or minimum) or possibly an inflection point.

First, let's find the derivative of h(t) with respect to t using the chain rule. The chain rule states that if we have a composite function f(g(t)), then its derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Let's start by finding the derivative of the inner function, g(t) = t^2 - 4:
g'(t) = 2t

Next, we find the derivative of the outer function, f(u) = u^(1/3):
f'(u) = (1/3)u^(-2/3)

Now, let's apply the chain rule:
h'(t) = f'(g(t)) * g'(t)
= (1/3)(t^2 - 4)^(-2/3)(2t)

Now we set the derivative equal to zero and solve for t to find the critical points:
(1/3)(t^2 - 4)^(-2/3)(2t) = 0

To solve this equation, we have two factors: (1/3), (t^2 - 4)^(-2/3), and 2t. For the equation to be equal to zero, at least one of these factors must be zero.

First, let's set (1/3) = 0. Since this is a constant, it can't be zero. So, we move on to the next factor.

Next, let's set (t^2 - 4)^(-2/3) = 0. This occurs when the expression inside the parentheses, t^2 - 4, is equal to zero:
t^2 - 4 = 0
t^2 = 4
t = ±2

Now, let's set 2t = 0:
2t = 0
t = 0

Therefore, the critical points of h(t) = (t^2-4)^(1/3) are t = -2, t = 0, and t = 2.