Question

let ( f(x)=(x - 3)^{-2} ). find all values of ( c ) in ( (1,7) ) such that ( f(7)-f(1)=f^{prime}(c)(7 - 1) ). (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 ( (1,7) ) such that ( f^{prime}(c)=\frac{f(7)-f(1)}{7 - 1} ).
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 ( (1,7) ), and there exists a ( c ) on ( (1,7) ) such that ( f^{prime}(c)=\frac{f(7)-f(1)}{7 - 1} ).
this contradicts the mean value theorem since there exists a ( c ) on ( (1,7) ) such that ( f^{prime}(c)=\frac{f(7)-f(1)}{7 - 1} ), but ( f ) is not continuous at ( x = 3 ).
nothing can be concluded.

Explanation:

Step1: Calculate \(f(7)\) and \(f(1)\)

Given \(f(x)=(x - 3)^{-2}=\frac{1}{(x - 3)^{2}}\), then \(f(7)=\frac{1}{(7 - 3)^{2}}=\frac{1}{16}\) and \(f(1)=\frac{1}{(1 - 3)^{2}}=\frac{1}{4}\).

Step2: Calculate \(f^{\prime}(x)\)

Using the power - rule \((u^n)^\prime=nu^{n - 1}u^\prime\), where \(u=x - 3\) and \(n=-2\). Then \(f^{\prime}(x)=-2(x - 3)^{-3}\times1=-\frac{2}{(x - 3)^{3}}\).

Step3: Substitute into the equation \(f(7)-f(1)=f^{\prime}(c)(7 - 1)\)

\(\frac{1}{16}-\frac{1}{4}=-\frac{2}{(c - 3)^{3}}\times6\).
First, simplify \(\frac{1}{16}-\frac{1}{4}=\frac{1 - 4}{16}=-\frac{3}{16}\).
The equation becomes \(-\frac{3}{16}=-\frac{12}{(c - 3)^{3}}\).
Cross - multiply: \(-3(c - 3)^{3}=-192\).
Divide both sides by \(-3\): \((c - 3)^{3}=64\).
Take the cube root of both sides: \(c-3 = 4\), so \(c = 7\). But \(c = 7
otin(1,7)\).

For the second part, the Mean Value Theorem requires that \(y = f(x)\) is continuous on the closed interval \([a,b]\) (in this case \([1,7]\)) and differentiable on the open interval \((a,b)\) (in this case \((1,7)\)). The function \(f(x)=\frac{1}{(x - 3)^{2}}\) has a discontinuity at \(x = 3\in[1,7]\) (since \(\lim_{x
ightarrow3}f(x)=\infty\)).

Answer:

\(c=\text{DNE}\)
For the conclusion: This does not contradict the Mean Value Theorem since \(f\) is not continuous at \(x = 3\).