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Question 7:

The problem asks:

1. Find the average or mean slope of the function $f(x) = \frac{1}{x}$ on the interval $[4, 7]$.
2. Find $c$ in the open interval $(4, 7)$ such that $f'(c)$ equals this mean slope, per the Mean Value Theorem (MVT).

Step 1: Calculate the mean slope

The formula for the mean slope is: $\text{Mean slope} = \frac{f(b) - f(a)}{b - a}$ Here, $a = 4$, $b = 7$, and $f(x) = \frac{1}{x}$.

$f(4) = \frac{1}{4} = 0.25, \quad f(7) = \frac{1}{7} \approx 0.142857$

$\text{Mean slope} = \frac{f(7) - f(4)}{7 - 4} = \frac{0.142857 - 0.25}{3} = \frac{-0.107143}{3} \approx -0.0357$

Step 2: Solve for $c$ using the Mean Value Theorem

The derivative of $f(x)$ is: $f'(x) = -\frac{1}{x^2}$

Set $f'(c) = \text{Mean slope}$: $-\frac{1}{c^2} = -0.0357$

Simplify: $\frac{1}{c^2} = 0.0357, \quad c^2 = \frac{1}{0.0357} \approx 28.035$

$c = \sqrt{28.035} \approx 5.29$

Thus, $c \approx 5.29$.

Question 8:

The problem asks:

1. Find the average or mean slope of the function $f(x) = 8\sqrt{x} + 2$ on the interval $[3, 10]$.

Step 1: Calculate the mean slope

The formula for the mean slope is the same: $\text{Mean slope} = \frac{f(b) - f(a)}{b - a}$

Here, $a = 3$, $b = 10$, and $f(x) = 8\sqrt{x} + 2$.

$f(3) = 8\sqrt{3} + 2 \approx 8(1.732) + 2 \approx 13.856$ $f(10) = 8\sqrt{10} + 2 \approx 8(3.162) + 2 \approx 27.296$

$\text{Mean slope} = \frac{f(10) - f(3)}{10 - 3} = \frac{27.296 - 13.856}{7} = \frac{13.44}{7} \approx 1.92$

Thus, the mean slope is approximately $1.92$.

Step 2: Find $c$ using the Mean Value Theorem

The derivative of $f(x) = 8\sqrt{x} + 2$ is: $f'(x) = \frac{8}{2\sqrt{x}} = \frac{4}{\sqrt{x}}$

Set $f'(c) = \text{Mean slope}$: $\frac{4}{\sqrt{c}} = 1.92$

Solve for $\sqrt{c}$: $\sqrt{c} = \frac{4}{1.92} \approx 2.083$

Square both sides: $c \approx (2.083)^2 \approx 4.34$

Thus, $c \approx 4.34$.

Summary of Results:

1. Question 7:
   • Mean slope: $-0.0357$
   • $c \approx 5.29$
2. Question 8:
   • Mean slope: $1.92$
   • $c \approx 4.34$

Let me know if you'd like a more detailed explanation or clarification!

Follow-up questions:

1. How do we interpret the Mean Value Theorem geometrically?
2. How does the derivative $f'(c)$ represent the mean slope on the interval?
3. Could there ever be more than one $c$ satisfying the Mean Value Theorem?
4. What happens if a function isn't differentiable on the interval?
5. Can the Mean Value Theorem apply to non-linear functions?

Tip: Always check the differentiability of the function over the interval when applying the Mean Value Theorem.