does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=ln (x)…

does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=ln (x), quad1,9$\nyes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.\nyes, $f$ is continuous on $1,9$ and differentiable on $(1,9)$.\nno, $f$ is not continuous on $1,9$.\nno, $f$ is continuous on $1,9$ but not differentiable on $(1,9)$.\nthere is not enough information to verify if this function satisfies the mean value theorem.\nif 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).

does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=ln (x), quad1,9$\nyes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.\nyes, $f$ is continuous on $1,9$ and differentiable on $(1,9)$.\nno, $f$ is not continuous on $1,9$.\nno, $f$ is continuous on $1,9$ but not differentiable on $(1,9)$.\nthere is not enough information to verify if this function satisfies the mean value theorem.\nif 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).

Answer

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)})