Derivative of Inverse Functions

[KingNothing]

let me start by saying that I'll be asking a lot of questions here soon as I'm preparing for a Calc BC AP test and I'm using practice exams to help.

On the first practice exam, the first question:
http://img390.imageshack.us/img390/9390/q1graph5bt.png

Find the derivative of $$f^{-1}(x)$$ at $$x=-.5$$.
Check my logic:

Since (1.5,.5) is on the original graph, (.5, 1.5) is on the inverse graph. Since the derivative of F at 1.5 is -.5, the answer is the recipricol, -2.

 

[AKG]

This isn't right. It's not clear why you did what you did to arrive at your answer though.

 

[KingNothing]

I'm using two properties of inverses:
If (a,b) is on the graph of F, (b,a) is on the graph of its inverse.
The slope of F at (a,b) is recipricol of the slope of its inverse at (b,a)

Therefore, since we want to find the slope of the inverse at (b,a), we need to look at the slope of the original at (a,b).

I know -2 is the correct answer, but since I have just learned the properties tonight, I am not 100% sure that my logic is correct.

 

[AKG]

The question you've asked asks for the derivative of f^-1 at x = 2. Nowhere in your work does "2" appear.

If x = 2, f^-1(x) = f^-1(2) = 0.5, so (2, 0.5) is on the graph of f^-1. To find the slope at (2, 0.5), you need to first find the slope of f at (0.5, 2), then take the reciprocal. The slope of f at (0.5, 2) is 4, so the slope of f^-1 at (2, 0.5) is 1/4.

You ask for the slope of f^-1 at 2, but your work gives its slope at 0.5

 

[KingNothing]

Sorry, I meant to type at $$x=-.5$$. A lot of confusion over nothing!

 

[VietDao29]

Sorry, I meant to type at $$x=-.5$$. A lot of confusion over nothing!

I must be missing something! But I didn't find any x value such that f(x) = -.5 in the table!
Are you sure it's not x = .5? If x = .5, then you seem to be correct. :)