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)…

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.)\nbased off of this information, what conclusions can be made about the mean value theorem?\nthis 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} ).\nthis does not contradict the mean value theorem since ( f ) is not continuous at ( x = 3 ).\nthis 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} ).\nthis 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 ).\nnothing can be concluded.

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.)\nbased off of this information, what conclusions can be made about the mean value theorem?\nthis 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} ).\nthis does not contradict the mean value theorem since ( f ) is not continuous at ( x = 3 ).\nthis 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} ).\nthis 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 ).\nnothing can be concluded.

Answer

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\notin(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\rightarrow3}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).