To find the derivative $g'(t)$ of the given function and determine the critical numbers, let's proceed step by step.

The function given is:

$g(t) = \frac{t}{\sqrt{8 - t}}$

Finding $g'(t)$

First, we can rewrite the function to make differentiation easier:

$g(t) = t (8 - t)^{-1/2}$

Now we use the product rule to find the derivative:

$g(t) = u \cdot v$ where: $u = t$ $v = (8 - t)^{-1/2}$

The product rule states that: $g'(t) = u'v + uv'$

First, we need to find the derivatives of $u$ and $v$:

$u' = \frac{d}{dt}(t) = 1$ $v = (8 - t)^{-1/2}$ $v' = \frac{d}{dt} (8 - t)^{-1/2}$

Using the chain rule for $v'$:

$v' = -\frac{1}{2} (8 - t)^{-3/2} \cdot (-1) = \frac{1}{2} (8 - t)^{-3/2}$

Now, applying the product rule:

$g'(t) = u'v + uv'$ $g'(t) = 1 \cdot (8 - t)^{-1/2} + t \cdot \frac{1}{2} (8 - t)^{-3/2}$ $g'(t) = (8 - t)^{-1/2} + \frac{t}{2} (8 - t)^{-3/2}$

Simplifying further:

[ g'(t) = \frac{1}{\sqrt{8 - t}} + \frac{t}{2(8 - t