Let's tackle this problem step by step. We'll first find the derivative $g'(t)$, then determine the values of $t$ where $g'(t) = 0$ or is undefined, and finally identify the critical numbers of the function.

1. Finding $g'(t)$

The function is given by: $g(t) = t \sqrt{8 - t}$ To find $g'(t)$, we can use the product rule of differentiation. The product rule states that if you have two functions $u(t)$ and $v(t)$, then: $(uv)' = u'v + uv'$

In our case: $u(t) = t$ $v(t) = \sqrt{8 - t}$

Let's find the derivatives $u'(t)$ and $v'(t)$: $u'(t) = 1$

For $v(t) = \sqrt{8 - t}$: $v(t) = (8 - t)^{1/2}$ Using the chain rule: $v'(t) = \frac{1}{2}(8 - t)^{-1/2} \cdot (-1) = -\frac{1}{2} (8 - t)^{-1/2} = -\frac{1}{2\sqrt{8 - t}}$

Now applying the product rule: $g'(t) = u'(t)v(t) + u(t)v'(t)$ $g'(t) = 1 \cdot \sqrt{8 - t} + t \cdot \left(-\frac{1}{2\sqrt{8 - t}}\right)$ $g'(t) = \sqrt{8 - t} - \frac{t}{2\sqrt{8 - t}}$ $g'(t) = \frac{2(8 - t) - t}{2\sqrt{8 - t}}$ $g'(t) = \frac{16 - 2t - t}{2\sqrt{8 - t}}$ $g'(t) = \frac{16 - 3t}{2\sqrt{8 - t}}$

2. Finding values of $c$ where $g'(c) = 0$ or undefined

To find where $g'(c) = 0$: $\frac{16 - 3c}{2\sqrt{8 - c}} = 0$ The numerator must be zero: $16 - 3c = 0$ $3c = 16$ $c = \frac{16}{3}$

To find where $g'(c)$ is undefined, we need to consider the denominator: $2\sqrt{8 - c} = 0$ $\sqrt{8 - c} = 0$ $8 - c = 0$ $c = 8$

However, $t < 7$ as given in the problem, so $c = 8$ is outside the domain of $g(t)$ and we exclude it.

So the values where $g'(c) = 0$ or undefined are: $c = \frac{16}{3}$

3. Critical numbers of the function

Critical numbers occur where $g'(t) = 0$ or $g'(t)$ is undefined. We have already found: $t = \frac{16}{3}$

Thus, the critical numbers of the function are: $t = \frac{16}{3}$

Summary

• Derivative: $g'(t) = \frac{16 - 3t}{2\sqrt{8 - t}}$
• Values of $c$ where $g'(c) = 0$ or undefined: $c = \frac{16}{3}$
• Critical numbers: $t = \frac{16}{3}$

Would you like any further details or explanations on this problem?

Here are 5 questions you can ask next:

1. Can you explain how the product rule works in differentiation?
2. How do you use the chain rule in differentiation?
3. What are the critical points of a function and why are they important?
4. How can you determine the domain of a function?
5. What are some common methods for finding the derivative of a function?

Tip: Always check the domain of the function before solving for critical points or values where the derivative is undefined to ensure you are considering only valid solutions.