etermine whether the Mean Value Theorem applies to the function \( f(x)=x+\frac{1}{x} \) on the interval \( [2,5] \) so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem.

To determine whether the **Mean Value Theorem (MVT)** applies to the function \( f(x) = x + \frac{1}{x} \) on the interval \([2, 5]\) and to find the point(s) guaranteed by the theorem, follow these steps: ### 1. Verify the Applicability of the Mean Value Theorem The **Mean Value Theorem** states that if a function \( f \) is: - **Continuous** on the closed interval \([a, b]\), and - **Differentiable** on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] **For \( f(x) = x + \frac{1}{x} \) on \([2, 5]\):** - **Continuity:** The function \( f(x) \) is continuous for all \( x > 0 \) since it is composed of continuous functions \( x \) and \( \frac{1}{x} \). Thus, \( f(x) \) is continuous on \([2, 5]\). - **Differentiability:** The function \( f(x) \) is differentiable for all \( x > 0 \) because both \( x \) and \( \frac{1}{x} \) are differentiable there. Thus, \( f(x) \) is differentiable on \((2, 5)\). **Conclusion:** The Mean Value Theorem **applies** to \( f(x) = x + \frac{1}{x} \) on the interval \([2, 5]\). ### 2. Find the Point \( c \) Guaranteed by the Mean Value Theorem **Step 1:** Calculate \( f(5) \) and \( f(2) \): \[ f(5) = 5 + \frac{1}{5} = 5.2 \] \[ f(2) = 2 + \frac{1}{2} = 2.5 \] **Step 2:** Compute the average rate of change over \([2, 5]\): \[ \frac{f(5) - f(2)}{5 - 2} = \frac{5.2 - 2.5}{3} = \frac{2.7}{3} = 0.9 \] **Step 3:** Find \( c \) such that \( f'(c) = 0.9 \). First, compute the derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx}\left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2} \] Set \( f'(c) = 0.9 \): \[ 1 - \frac{1}{c^2} = 0.9 \] \[ \frac{1}{c^2} = 1 - 0.9 = 0.1 \] \[ c^2 = \frac{1}{0.1} = 10 \] \[ c = \sqrt{10} \approx 3.1623 \] **Conclusion:** There exists a point \( c = \sqrt{10} \) (approximately \( 3.1623 \)) in the interval \((2, 5)\) where: \[ f'(c) = 0.9 = \frac{f(5) - f(2)}{5 - 2} \] This point \( c \) is the one guaranteed by the Mean Value Theorem for the given function and interval. **Final Answer:** Yes, the Mean Value Theorem applies. The required point is c = √10, which is approximately 3.162, in the interval [2, 5].