Problem 7: Let $$ f(x)=\sqrt{x} $$ on the interval $$ [1,9] $$. - Use the MVT to find the value(s) of $$ c $$ in $$ (1,9) $$ where $$ f^{\prime}(c)=\frac{f(9)-f(1)}{9-1} $$

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To apply the **Mean Value Theorem (MVT)** to the function $$ f(x) = \sqrt{x} $$ on the interval $$[1, 9]$$, follow these steps:

### **Step 1: Verify the Conditions for MVT**

The Mean Value Theorem states that if a function $$ f $$ is:
1. **Continuous** on the closed interval $$[a, b]$$, and
2. **Differentiable** on the open interval $$ (a, b) $$,

then there exists at least one $$ c $$ in $$ (a, b) $$ such that:

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

**For $$ f(x) = \sqrt{x} $$:**
- **Continuity:** $$ \sqrt{x} $$ is continuous for all $$ x \geq 0 $$, so it's continuous on $$[1, 9]$$.
- **Differentiability:** $$ \sqrt{x} $$ is differentiable for all $$ x > 0 $$, so it's differentiable on $$ (1, 9) $$.

Both conditions are satisfied, so the MVT applies.

### **Step 2: Compute $$ f(a) $$ and $$ f(b) $$**

Given:
- $$ a = 1 $$
- $$ b = 9 $$

$$
f(a) = f(1) = \sqrt{1} = 1
$$
$$
f(b) = f(9) = \sqrt{9} = 3
$$

### **Step 3: Calculate the Average Rate of Change**

$$
\frac{f(b) - f(a)}{b - a} = \frac{3 - 1}{9 - 1} = \frac{2}{8} = \frac{1}{4}
$$

### **Step 4: Find $$ f'(x) $$**

First, find the derivative of $$ f(x) = \sqrt{x} $$:

$$
f(x) = x^{1/2} \implies f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
$$

### **Step 5: Set $$ f'(c) $$ Equal to the Average Rate and Solve for $$ c $$**

$$
f'(c) = \frac{1}{2\sqrt{c}} = \frac{1}{4}
$$

Solve for $$ c $$:

$$
\frac{1}{2\sqrt{c}} = \frac{1}{4}
$$
$$
2\sqrt{c} = 4 \quad \text{(cross-multiplying)}
$$
$$
\sqrt{c} = 2 \quad \text{(dividing both sides by 2)}
$$
$$
c = 4 \quad \text{(squaring both sides)}
$$

### **Conclusion**

The value of $$ c $$ in $$ (1, 9) $$ that satisfies the Mean Value Theorem for $$ f(x) = \sqrt{x} $$ on the interval $$[1, 9]$$ is:

$$
\boxed{4}
$$
The value of $$ c $$ is 4.

Problem 7:
Let on the interval .

Use the MVT to find the value(s) of in where

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Problem 7: Let on the interval . Use the MVT to find the value(s) of in where