Consider the function f ( x )

Step 1: Find the Average Slope

The average slope of a function $f(x)$ over the interval $[a, b]$ is given by:

$\text{Average Slope} = \frac{f(b) - f(a)}{b - a}$

Here, $f(x) = 4\sqrt{x} + 5$, and the interval is $[1, 8]$.

1. Evaluate $f(1)$: $f(1) = 4\sqrt{1} + 5 = 4(1) + 5 = 9$

2. Evaluate $f(8)$: $f(8) = 4\sqrt{8} + 5 = 4(2\sqrt{2}) + 5 = 8\sqrt{2} + 5$

3. Compute the average slope: $\text{Average Slope} = \frac{f(8) - f(1)}{8 - 1} = \frac{(8\sqrt{2} + 5) - 9}{7} = \frac{8\sqrt{2} - 4}{7}$

Thus, the average slope is: $\text{Average Slope} = \frac{8\sqrt{2} - 4}{7}$

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Step 2: Use the Mean Value Theorem to Find $c$

The Mean Value Theorem guarantees a $c$ in $(1, 8)$ such that:

$f'(c) = \text{Average Slope}$

Find $f'(x)$:

The derivative of $f(x) = 4\sqrt{x} + 5$ is: $f'(x) = \frac{4}{2\sqrt{x}} = \frac{2}{\sqrt{x}}$

Set $f'(c) = \frac{8\sqrt{2} - 4}{7}$: $\frac{2}{\sqrt{c}} = \frac{8\sqrt{2} - 4}{7}$

Solve for $c$:

1. Multiply both sides by $\sqrt{c}$: $2 = \sqrt{c} \cdot \frac{8\sqrt{2} - 4}{7}$

2. Solve for $\sqrt{c}$: $\sqrt{c} = \frac{2 \cdot 7}{8\sqrt{2} - 4} = \frac{14}{8\sqrt{2} - 4}$

3. Square both sides to find $c$: $c = \left(\frac{14}{8\sqrt{2} - 4}\right)^2$

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Final Results:

1. Average slope: $\frac{8\sqrt{2} - 4}{7}$
2. Value of $c$: $\left(\frac{14}{8\sqrt{2} - 4}\right)^2$ (simplify further if needed).

Would you like more explanation or further simplification?

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Follow-Up Questions:

1. What does the Mean Value Theorem guarantee about continuous functions?
2. How does $f'(x) = \frac{2}{\sqrt{x}}$ change as $x$ increases?
3. What is the geometric meaning of the average slope in this context?
4. Can you confirm the existence of $c$ without solving explicitly?
5. How does the Mean Value Theorem relate to the Intermediate Value Theorem?

Tip:

When working with square roots, it’s often helpful to rationalize the denominator or simplify expressions to check for numerical or algebraic errors.