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 to the function.

Explanation:

Step1: Check continuity and differentiability

The function \(y = \ln(17x)\) is a logarithmic function. The domain of \(y=\ln(17x)\) is \(x>0\). Since the interval \([1,e]\) (\(1 > 0\) and \(e>0\)) and the function \(y = \ln(17x)\) is continuous for all \(x>0\) and differentiable for all \(x > 0\) (the derivative \(y^\prime=\frac{1}{x}\) exists for \(x>0\)). So the Mean - Value Theorem applies.

Step2: Use the Mean - Value Theorem formula

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(f^\prime(c)=\frac{f(b)-f(a)}{b - a}\).
Here, \(a = 1\), \(b=e\), \(f(x)=\ln(17x)\), \(f(1)=\ln(17\times1)=\ln(17)\), \(f(e)=\ln(17e)=\ln(17)+\ln(e)=\ln(17)+1\).
Then \(\frac{f(e)-f(1)}{e - 1}=\frac{(\ln(17)+1)-\ln(17)}{e - 1}=\frac{1}{e - 1}\).
Also, \(f^\prime(x)=\frac{1}{x}\), setting \(f^\prime(c)=\frac{1}{c}\) equal to \(\frac{1}{e - 1}\).

Step3: Solve for \(c\)

We have the equation \(\frac{1}{c}=\frac{1}{e - 1}\), cross - multiplying gives \(c=e - 1\approx1.718\) (since \(e\approx2.718\)) and \(1

Answer:

a. The Mean Value Theorem applies to the function.
b. The point \(c = e-1\) (where \(e\approx2.718\), so \(c\approx1.718\)) is guaranteed to exist by the Mean Value Theorem.