Estimated Derivative of Inverse Function at x=1

[UrbanXrisis]

f(x) is 1.8 when x=1
f(x) is 0.5 when x=2.8
f(x) is 1.5 when x=1.2

what is the best estimate for the value of (f^-1)'1?

I now that f'(1) is -1.6 because of linear apporximation:
(2.8-1.2)/(.5-1.5)=1.6/-1=-1.6

so then (f^-1)'1 = 1/1.6?

 

[ponjavic]

How do you know that f(x) is linear?

 

[UrbanXrisis]

sorry, that method is "local approximation", not linear approximation

 

[HallsofIvy]

f(x) is 1.8 when x=1
f(x) is 0.5 when x=2.8
f(x) is 1.5 when x=1.2

what is the best estimate for the value of (f^-1)'1?

I now that f'(1) is -1.6 because of linear apporximation:
(2.8-1.2)/(.5-1.5)=1.6/-1=-1.6

so then (f^-1)'1 = 1/1.6?

Your confusing f(x) with f^-1(x) so you have your formula "upside down".
I would interpret this as saying that f^-1(0.5)= 2.8, f^-1(1.0)= 1.8, and f^-1(1.5)= 1.2. That's clearly a decreasing function
You are given three equally spaced points and a good estimate for the derivative of a function, in that situation, is (f(a+h)- f(a-h))/(2h). In terms of the inverse function, we have a= 1, h= 0.5 so that
f^-1(1- 0.5)= f^-1(0.5)= 2.8 and f^-1(1+ 0.5)= f^-1(1.5)= 1.2.
The estimate for the derivative of f^-1at x=1 (not f' (1))
is (1.2- 2.8)/(2(0.5))= -1.6. That is the derivative of f^-1(x) at x= 1, not of f(x).