If the function is [tex](x-3)^{-2}[/tex] and f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.
Given function is [tex](x-3)^{-2}[/tex] 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)=[tex](2-3)^{-2}[/tex]
f(2)=1
f(5)=[tex](5-3)^{-2}[/tex]
f(5)=1
And the slope of the function:
[tex]f^{1}[/tex](x)=[tex]f(5)-f(2)/(5-2)[/tex]
[tex]f^{1}[/tex](c)=0
Now,
[tex]f^{1} (x)=-2*(x-3)^{-3}[/tex]
=-2[tex](x-3)^{-3}[/tex]
=0
-2 is not equal to 0. So there is not any answer.
Hence if the function is [tex](x-3)^{-2}[/tex] and f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.
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