Problem 10: Let $$ f(x)=\frac{1}{x} $$ on the interval $$ [1,4] $$. - Use the MVT to find the value(s) of $$ c $$ in $$ (1,4) $$ where $$ f^{\prime}(c)=\frac{f(4)-f(1)}{4-1} $$

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To apply the Mean Value Theorem (MVT) to the function $$ f(x) = \frac{1}{x} $$ on the interval $$[1, 4]$$, follow these steps:

1. **Verify the Conditions for MVT:**
   - **Continuity:** $$ f(x) = \frac{1}{x} $$ is continuous on $$[1, 4]$$.
   - **Differentiability:** $$ f(x) = \frac{1}{x} $$ is differentiable on $$(1, 4)$$.

Since both conditions are satisfied, MVT applies.

2. **State the MVT:**
   There exists at least one $$ c $$ in $$(1, 4)$$ such that:
   $$
   f'(c) = \frac{f(4) - f(1)}{4 - 1}
   $$

3. **Compute $$ f(4) $$ and $$ f(1) $$:**
   $$
   f(4) = \frac{1}{4}, \quad f(1) = 1
   $$

4. **Calculate the Right-Hand Side of MVT:**
   $$
   \frac{f(4) - f(1)}{4 - 1} = \frac{\frac{1}{4} - 1}{3} = \frac{-\frac{3}{4}}{3} = -\frac{1}{4}
   $$

5. **Find $$ f'(x) $$:**
   $$
   f'(x) = -\frac{1}{x^2}
   $$

6. **Set $$ f'(c) $$ Equal to $$-\frac{1}{4}$$:**
   $$
   -\frac{1}{c^2} = -\frac{1}{4} \implies \frac{1}{c^2} = \frac{1}{4}
   $$

7. **Solve for $$ c $$:**
   $$
   c^2 = 4 \implies c = 2 \quad (\text{since } c \in (1, 4))
   $$

**Conclusion:**
The value of $$ c $$ that satisfies the Mean Value Theorem for the given function on $$[1, 4]$$ is:

$$\boxed{2}$$
The value of $$ c $$ is 2.

Problem 10: Let \( f(x)=\frac{1}{x} \) on the interval \( [1,4] \). - Use the MVT to find the value(s) of \( c \) in \( (1,4) \) where \( f^{\prime}(c)=\frac{f(4)-f(1)}{4-1} \)

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