The problem asks to find the value or values of $c$ that satisfy the equation

$\frac{f(b) - f(a)}{b - a} = f'(c)$

for the given function and interval, using the Mean Value Theorem (MVT).

Step 1: Analyze the Given Function and Interval

The function is: $f(x) = 7x + \frac{7}{x}$ And the interval is $\left[\frac{1}{10}, 10\right]$.

Step 2: Mean Value Theorem

According to the Mean Value Theorem, there exists at least one value $c$ in the open interval $\left( \frac{1}{10}, 10 \right)$ such that:

$f'(c) = \frac{f(10) - f\left(\frac{1}{10}\right)}{10 - \frac{1}{10}}$

Step 3: Compute $f'(x)$

To apply the MVT, we first need the derivative $f'(x)$.

$f(x) = 7x + \frac{7}{x}$ Using the power rule and the chain rule, we differentiate:

$f'(x) = 7 - \frac{7}{x^2}$

Step 4: Calculate $f(10)$ and $f\left(\frac{1}{10}\right)$

• $f(10) = 7(10) + \frac{7}{10} = 70 + 0.7 = 70.7$
• $f\left(\frac{1}{10}\right) = 7\left(\frac{1}{10}\right) + \frac{7}{\frac{1}{10}} = 0.7 + 70 = 70.7$

Step 5: Compute the Difference Quotient

Now calculate:

$\frac{f(10) - f\left(\frac{1}{10}\right)}{10 - \frac{1}{10}} = \frac{70.7 - 70.7}{10 - 0.1} = \frac{0}{9.9} = 0$

Thus, the right-hand side of the Mean Value Theorem equation is $0$.

Step 6: Set $f'(c) = 0$

From the MVT, we now have:

$7 - \frac{7}{c^2} = 0$

Solve for $c$:

$7 = \frac{7}{c^2}$

$c^2 = 1$

$c = \pm 1$

Step 7: Check If $c$ Lies in the Interval

The interval is $\left(\frac{1}{10}, 10\right)$. So, $c = 1$ is the only valid solution because $c = -1$ is not in the interval.

Final Answer:

The value of $c$ that satisfies the Mean Value Theorem is:

$c = 1$

Would you like more details or have any other questions?

Here are 5 relative questions for further exploration:

1. What happens if the function was different, such as $f(x) = x^3$?
2. How would the problem change if the interval were $[2, 5]$?
3. Why does the MVT require a differentiable function on an open interval?
4. How is the Mean Value Theorem applied in real-world scenarios?
5. What is the difference between the Mean Value Theorem and Rolle's Theorem?

Tip: The Mean Value Theorem guarantees at least one value $c$, but in some cases, multiple $c$-values might satisfy the equation depending on the function's behavior.