Question

does the function satisfy the hypotheses of the mean value theorem on the given interval?
$f(x)=ln (x), quad1,9$
yes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.
yes, $f$ is continuous on $1,9$ and differentiable on $(1,9)$.
no, $f$ is not continuous on $1,9$.
no, $f$ is continuous on $1,9$ but not differentiable on $(1,9)$.
there is not enough information to verify if this function satisfies the mean value theorem.
if it satisfies the hypotheses, find all numbers $c$ that satisfy the conclusion of the mean value theorem. (enter your answers as a comma-separated list. if it does not satisfy the hypotheses, enter dne).

Explanation:

Step1: Check continuity and differentiability

The function \(y = \ln(x)\) is continuous for \(x>0\) and differentiable for \(x > 0\). On the interval \([1,9]\), \(f(x)=\ln(x)\) is continuous on \([1,9]\) (since \(1>0\)) and differentiable on \((1,9)\) (because the derivative \(f^{\prime}(x)=\frac{1}{x}\) exists for \(x\in(1,9)\)).

Step2: Apply the Mean - Value Theorem formula

The Mean - Value Theorem states that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\), where \(a = 1\), \(b=9\), \(f(x)=\ln(x)\), \(f(a)=\ln(1) = 0\), \(f(b)=\ln(9)\), and \(f^{\prime}(x)=\frac{1}{x}\).
Substitute into the formula: \(\frac{1}{c}=\frac{\ln(9)-\ln(1)}{9 - 1}\).
Since \(\ln(1) = 0\), we have \(\frac{1}{c}=\frac{\ln(9)}{8}\).
We know that \(\ln(9)=2\ln(3)\), so \(\frac{1}{c}=\frac{2\ln(3)}{8}=\frac{\ln(3)}{4}\).
Then \(c=\frac{4}{\ln(3)}\approx\frac{4}{1.0986}\approx3.64\).

Answer:

B. Yes, \(f\) is continuous on \([1,9]\) and differentiable on \((1,9)\); \(c = \frac{4}{\ln(3)}\)