12. Let \( f(x)=x^{2}+\sqrt[3]{x} \). Find the value(s) which satisfy the Mean Value Theorem, \( f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \), on the interval \( [1,8] \)

To find the value(s) \( c \) that satisfy the Mean Value Theorem for the function \( f(x) = x^2 + \sqrt[3]{x} \) on the interval \([1, 8]\), follow these steps: ### Step 1: Apply the Mean Value Theorem (MVT) The Mean Value Theorem states that there exists at least one \( c \) in the open interval \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Here, \( a = 1 \) and \( b = 8 \). ### Step 2: Calculate \( f(b) \) and \( f(a) \) \[ f(8) = 8^2 + \sqrt[3]{8} = 64 + 2 = 66 \] \[ f(1) = 1^2 + \sqrt[3]{1} = 1 + 1 = 2 \] \[ \frac{f(8) - f(1)}{8 - 1} = \frac{66 - 2}{7} = \frac{64}{7} \approx 9.142857 \] ### Step 3: Find the Derivative \( f'(x) \) \[ f'(x) = 2x + \frac{1}{3x^{2/3}} \] ### Step 4: Set Up the Equation Set the derivative equal to the average rate of change: \[ 2c + \frac{1}{3c^{2/3}} = \frac{64}{7} \] ### Step 5: Solve for \( c \) This equation is transcendental and doesn't easily solve algebraically, so we'll use numerical methods (like the Newton-Raphson method) to approximate \( c \). Through iterative approximation: - **Initial Guess**: \( c = 4.5 \) - **Calculation**: \[ f(4.5) \approx 2(4.5) + \frac{1}{3(4.5)^{2/3}} \approx 9 + 0.123 \approx 9.123 \] Since \( 9.123 \) is very close to \( 9.142857 \), we refine our approximation further: - **Refined Value**: \( c \approx 4.51 \) ### Conclusion The value of \( c \) that satisfies the Mean Value Theorem for the given function on the interval \([1, 8]\) is approximately: \[ c \approx 4.51 \] **Answer:** After simplifying, the solution c is approximately 4.51.