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 \( f^{\prime}(c)=\frac{f(9)-f(1)}{9-1} \)

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: \[ \boxed{4} \]