Consider the following function.
g(t) = t

sqrt8 − t
,    t < 7
Find
g'(t).
g'(t) =

Excluding any values not in the domain of g, for what values of c is
g'(c) = 0
or undefined? (Enter your answers as a comma-separated list.)
c =

Give the critical numbers of the function. (Enter your answers as a comma-separated list.)
t =

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.

Learn how to find the derivative and critical numbers of the function g(t) = t * sqrt(8 - t). Detailed step-by-step solution with application of differentiation rules. Suitable for college-level mathematics.

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