Question

attempt 1: 2 attempts remaining. consider the function f(x)=6 - 2/x over the interval 1,11. find the value(s) c that satisfies the conclusion of the mean value theorem. if there is more than one value, enter them as a comma - separated list.

Explanation:

Step1: Recall Mean - Value Theorem formula

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\), where \(a = 1\), \(b = 11\) and \(f(x)=6-\frac{2}{x}\). First, find \(f(1)\) and \(f(11)\).
\[f(1)=6 - 2=4\]
\[f(11)=6-\frac{2}{11}=\frac{66 - 2}{11}=\frac{64}{11}\]
Then \(\frac{f(11)-f(1)}{11 - 1}=\frac{\frac{64}{11}-4}{10}=\frac{\frac{64 - 44}{11}}{10}=\frac{\frac{20}{11}}{10}=\frac{2}{11}\)

Step2: Find the derivative of \(f(x)\)

Differentiate \(f(x)=6-\frac{2}{x}=6-2x^{-1}\) using the power - rule \((x^n)^\prime=nx^{n - 1}\). So \(f^{\prime}(x)=2x^{-2}=\frac{2}{x^{2}}\)

Step3: Solve for \(c\)

Set \(f^{\prime}(c)=\frac{2}{11}\), so \(\frac{2}{c^{2}}=\frac{2}{11}\). Cross - multiply to get \(2\times11 = 2c^{2}\). Divide both sides by 2: \(c^{2}=11\). Since \(c\in(1,11)\), then \(c=\sqrt{11}\)

Answer:

\(\sqrt{11}\)