Question

Let f(x) = (x − 3)−2. find all values of c in (1, 4) such that f(4) − f(1) = f '(c)(4 − 1). (enter your answers as a comma-separated list. if an answer does not exist, enter dne. ) c =

Answer 1

If the function is $(x-3)^{-2}$ and  f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.

Given function is $(x-3)^{-2}$ and  f(4) − f(1) = f '(c)(4 − 1).

In this question we have to apply the mean value theorem, which says that given a secant line between points A and B, there is at least a point C that belongs to the curve and the derivative of that curve exists.

We begin by calculating f(2) and f(5):

f(2)=$(2-3)^{-2}$

f(2)=1

f(5)=$(5-3)^{-2}$

f(5)=1

And the slope of the function:

$f^{1}$(x)=$f(5)-f(2)/(5-2)$

$f^{1}$(c)=0

Now,

$f^{1} (x)=-2*(x-3)^{-3}$

=-2$(x-3)^{-3}$

=0

-2 is not equal to 0. So there is not any answer.

Hence if the function is $(x-3)^{-2}$ and  f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.

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