Question

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\ln x, \quad g(x)=x^{3} \end{array} \quad \frac{\text { Interval }}{\left[1,4\right]}\)

Short answer

The value \( c \) within the interval [1,4] that satisfies the extension of Mean Value Theorem for the functions \( f(x) = \ln x \) and \( g(x) = x^3 \) is \( c = \sqrt[3]{\frac{63}{3 \ln 4}} \).

Step-by-step solution

Understanding the Expression

Here, we are given two functions \( f(x) = \ln x \) and \( g(x) = x^3 \) and an interval [1,4]. The expression \( \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \) is the Extended Mean Value Theorem.

Compute the Derivatives of the Functions

The derivative of \( f(x) = \ln x \) is \( f'(x) = \frac{1}{x} \). The derivative of \( g(x) = x^3 \) is \( g'(x) = 3x^2 \).

Substitute the Given Functions into the Expression

Set up the expression to find \( c \). \nSo, \(\frac{f'(c)}{g'(c)} = \frac{f(4) - f(1)}{g(4) - g(1)} \) becomes \( \frac{\frac{1}{c}}{3c^2} = \frac{\ln 4 - \ln 1}{4^3 - 1^3} \). Simplifying right side we have \( \frac{\ln 4}{63} \).

Simplify and Solve for \( c \)

Cross multiply and solve for \( c \): \( \frac{1}{c} = \frac{3c^2 \ln 4}{63} \). Solving this equation for \( c \), we get \( c^3 = \frac{63}{3 \ln 4} \), so \( c = \sqrt[3]{\frac{63}{3 \ln 4}} \). We can confirm that this solution lies within the interval [1,4].