Question

a. determine whether the mean value theorem applies to the function ( f(x)=ln 17x ) on the given interval (1,e).
b. if so, find or approximate the point(s) that are guaranteed to exist by the mean value theorem.
a. the mean value theorem applies to the function.
b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the point(s) is/are ( x=)
(type an exact answer. use a comma to separate answers as needed.)
b. the mean value theorem does not apply in this case.

Explanation:

Step1: Recall the 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 there exists at least one \(c\in(a,b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\).
For \(f(x)=\ln(17x)\), its derivative \(f^{\prime}(x)=\frac{1}{x}\) (using the chain - rule: if \(y=\ln(u)\) and \(u = 17x\), then \(y^{\prime}=\frac{u^{\prime}}{u}=\frac{17}{17x}=\frac{1}{x}\)). \(f(x)\) is continuous on \([1,e]\) (since the domain of \(y = \ln(17x)\) is \(x>0\)) and differentiable on \((1,e)\).

Step2: Calculate \(f(1)\) and \(f(e)\)

\(f(1)=\ln(17\times1)=\ln(17)\)
\(f(e)=\ln(17e)=\ln(17)+\ln(e)=\ln(17) + 1\)
Then \(\frac{f(e)-f(1)}{e - 1}=\frac{(\ln(17)+1)-\ln(17)}{e - 1}=\frac{1}{e - 1}\)

Step3: Solve \(f^{\prime}(c)=\frac{f(e)-f(1)}{e - 1}\)

Set \(f^{\prime}(c)=\frac{1}{c}\) equal to \(\frac{1}{e - 1}\).
We get the equation \(\frac{1}{c}=\frac{1}{e - 1}\)
Cross - multiply: \(c=e - 1\)

Answer:

A. The point(s) is/are \(x = e-1\)