Question

let (f(x)=(x - 3)^{-2}). find all values of (c) in ((2,5)) such that (f(5)-f(2)=f(c)(5 - 2)). (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

based off of this information, what conclusions can be made about the mean value theorem?

this contradicts the mean value theorem since (f) satisfies the hypotheses on the given interval but there does not exist any (c) on ((2,5)) such that (f(c)=\frac{f(5)-f(2)}{5 - 2}).

this does not contradict the mean value theorem since (f) is not continuous at (x = 3).

this does not contradict the mean value theorem since (f) is continuous on ((2,5)), and there exists a (c) on ((2,5)) such that (f(c)=\frac{f(5)-f(2)}{5 - 2}).

this contradicts the mean value theorem since there exists a (c) on ((2,5)) such that (f(c)=\frac{f(5)-f(2)}{5 - 2}), but (f) is not continuous at (x = 3).

nothing can be concluded.

Explanation:

Step1: Calculate \(f(5)\) and \(f(2)\)

First, find \(f(x)=(x - 3)^{-2}=\frac{1}{(x - 3)^2}\). Then \(f(5)=\frac{1}{(5 - 3)^2}=\frac{1}{4}\), and \(f(2)=\frac{1}{(2 - 3)^2}=1\).

Step2: Calculate \(\frac{f(5)-f(2)}{5 - 2}\)

\(\frac{f(5)-f(2)}{5 - 2}=\frac{\frac{1}{4}-1}{3}=\frac{\frac{1 - 4}{4}}{3}=\frac{-\frac{3}{4}}{3}=-\frac{1}{4}\).

Step3: Find the derivative \(f'(x)\)

Using the power - rule and chain - rule, if \(y=(x - 3)^{-2}\), let \(u=x - 3\), then \(y = u^{-2}\). \(\frac{dy}{du}=-2u^{-3}\) and \(\frac{du}{dx}=1\), so \(f'(x)=-2(x - 3)^{-3}=-\frac{2}{(x - 3)^3}\).

Step4: Solve for \(c\)

Set \(f'(c)=-\frac{1}{4}\), so \(-\frac{2}{(c - 3)^3}=-\frac{1}{4}\). Cross - multiply to get \((c - 3)^3 = 8\). Then \(c-3 = 2\), and \(c=5\). But \(c\in(2,5)\), so \(c = DNE\).

Step5: Analyze the Mean Value Theorem

The function \(f(x)=\frac{1}{(x - 3)^2}\) is not continuous at \(x = 3\in(2,5)\). The Mean Value Theorem requires the function to be continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\). Since \(f(x)\) is not continuous at \(x = 3\), this does not contradict the Mean Value Theorem.

Answer:

DNE
This does not contradict the Mean Value Theorem since \(f\) is not continuous at \(x = 3\).