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Find all values of \( c \) on the interval \( [a, b] \) that satisfy the conclusion of the Mean Value Theorem. (2) \( f(x)=x^{2}+x \) on \( [2,4] \) \( f^{\prime}(x) 2 x-1=\frac{f(y)-f(2)}{4-2} \frac{20-6}{6}=\frac{14}{6} \) \[ x=\frac{4}{3} \] \( x=\frac{4}{3}-\frac{2}{3} \)

Ask by Tucker Campos. in the United States
Feb 05,2025

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Tutor-Verified Answer

Answer

The value \( c = 3 \) satisfies the Mean Value Theorem on the interval \([2,\,4]\).

Solution

To apply the **Mean Value Theorem (MVT)** to the function \( f(x) = x^2 + x \) on the interval \([2, 4]\), let's follow the correct steps: ### Mean Value Theorem Statement If \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] ### Step-by-Step Application 1. **Verify Continuity and Differentiability:** - \( f(x) = x^2 + x \) is a polynomial, which is continuous and differentiable everywhere. Thus, it satisfies the conditions of the MVT on \([2, 4]\). 2. **Compute \( f(a) \) and \( f(b) \):** \[ f(2) = 2^2 + 2 = 4 + 2 = 6 \] \[ f(4) = 4^2 + 4 = 16 + 4 = 20 \] 3. **Calculate the Average Rate of Change:** \[ \frac{f(4) - f(2)}{4 - 2} = \frac{20 - 6}{2} = \frac{14}{2} = 7 \] 4. **Find \( f'(x) \):** \[ f'(x) = \frac{d}{dx}(x^2 + x) = 2x + 1 \] 5. **Set \( f'(c) = 7 \) and Solve for \( c \):** \[ 2c + 1 = 7 \] \[ 2c = 6 \] \[ c = 3 \] ### Conclusion There is **one value** of \( c \) in the interval \((2, 4)\) that satisfies the conclusion of the Mean Value Theorem: \[ c = 3 \] This means that at \( x = 3 \), the instantaneous rate of change of \( f(x) \) equals the average rate of change over \([2, 4]\). **Final Answer:** The value \( c = 3 \) satisfies the Mean Value Theorem on the interval \([2,\,4]\).

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To apply the Mean Value Theorem (MVT) effectively, first, we confirm that the function \( f(x) = x^2 + x \) is continuous on the closed interval \([2, 4]\) and differentiable on the open interval \((2, 4)\). Now, let's find \( f(2) \) and \( f(4) \): \[ f(2) = 2^2 + 2 = 4 + 2 = 6 \] \[ f(4) = 4^2 + 4 = 16 + 4 = 20 \] Next, we compute the average rate of change of \( f \) from \( x=2 \) to \( x=4 \): \[ \text{Average Rate of Change} = \frac{f(4) - f(2)}{4 - 2} = \frac{20 - 6}{2} = \frac{14}{2} = 7 \] According to the MVT, there exists at least one \( c \) in \((2, 4)\) such that \( f'(c) = 7 \). Now, let's compute the derivative \( f'(x) \): \[ f'(x) = 2x + 1 \] We set this equal to the average rate of change we calculated: \[ 2c + 1 = 7 \] Now, solve for \( c \): \[ 2c = 7 - 1 = 6 \quad \Rightarrow \quad c = \frac{6}{2} = 3 \] So, the value of \( c \) that satisfies the conclusion of the Mean Value Theorem in the interval \([2, 4]\) is \( c = 3 \). Did you catch that? We confirmed \( c = 3\) is not only in our interval but also solidifies the funky math of calculus! 🌟✨ Now, let's dig deeper into how such fun functions have shaped calculations over time! Historically, the roots of calculus were laid by mathematicians like Newton and Leibniz in the late 17th century when they developed methods to solve problems involving rates of change and areas under curves. The Mean Value Theorem itself, introduced by Cauchy in the 19th century, provides crucial insights connecting derivatives and slopes of secant lines, paving the way for more advanced calculus applications! In the real world, the Mean Value Theorem is like a secret sauce in various fields! Whether you’re optimizing a delivery route or analyzing how quickly a car accelerates, this theorem helps identify that crucial moment when instantaneous change matches average change. It’s not just mathematics; it’s about understanding motion and flow in our everyday lives! 🚗💨

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