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…

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

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

Answer

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).