# Derivative calc problem

1. Apr 15, 2009

### coookiemonste

1. The problem statement, all variables and given/known data
If f(x)=x^3-x-6 and g is the inverse function of f.
what is f'(g(0))g'(0)?
the answer is 1...but im not sure how to get that.

2. Relevant equations
[f(g(x))]'=f'g(x)*g'(x)

3. The attempt at a solution
i wasnt quite sure, but i assumed g(x)=1/(x^3-x-6)
^is this assumption right?
then f'(x)=3x^2-1
i also set f(x)=0 to find x=2.
but for g(0) i get -1/6
so for f'(g(0))g'(0)= i get -1(-1/6)*(-1) = -1/6
but this is incorrect.

2. Apr 15, 2009

### meiso

An inverse function can be found by switching the positions of the x and y variables of a function and (to put it in true function form) solving for y.

For example, if we have

$$f(x) = y = 3x -1$$

and we want to find the inverse of f, which we can call g, these are the steps to take:

1. Switch the positions of the x and y variables:

$$x = 3y - 1$$

2. Solve for y:

$$x + 1 = 3y$$

$$\frac{1}{3}x + \frac{1}{3} = y$$

So we have,

$$g(x) = \frac{1}{3}x + \frac{1}{3}$$

Now, watch what happens if we take f(g(x)):

$$f(g(x)) = f(\frac{1}{3}x + \frac{1}{3})$$

$$= 3(\frac{1}{3}x + \frac{1}{3}) - 1$$

$$= x + 1 - 1$$

$$= x$$

Similarly, if we try g(f(x)):

$$g(f(x)) = g(3x -1)$$

$$= \frac{1}{3}(3x -1) + \frac{1}{3}$$

$$= x - \frac{1}{3} + \frac{1}{3}$$

$$= x$$

This happens with any function and its true inverse. If we take the composite of a function and its inverse in either directions (i.e. f(g(x)) of g(f(x)) ), we end up getting x back. Or in the case of a number, e.g. f(g(1)) = g(f(1)) = 1

An inverse function basically "undoes" whatever a function does to a number.

So, your g(x) is not correct. And you may not always be able to easily find the inverse of a function solved for y, as is true with the function you have been given.

That being said, there are a couple ways to solve your problem.

The quickest way is to try playing around with the formula you gave under "relevant equations," to see if any expression simplifies, then take it from there.

Otherwise, you can look up a formula for the derivative of an inverse function to find g'(x) by skipping right to the derivative of an inverse function without finding the inverse. It can be found in any calculus textbook.

3. Apr 15, 2009

### Staff: Mentor

In case it wasn't abundantly clear from what Meiso said, the inverse of f(x) is not 1/(x^3 -x - 6). This is the reciprocal of f(x), but not its inverse.

4. Apr 15, 2009

### coookiemonste

so if f(g(x))=x
and [f(g(x))]'=(x)'=1
is this correct?

5. Apr 15, 2009

### Staff: Mentor

Yes.

6. Apr 15, 2009

### meiso

Exactly right. And, sorry, I thought I was clear enough when I said,

Last edited: Apr 15, 2009
7. Apr 15, 2009

### Staff: Mentor

You were clear, but what you said was a little hard to find, being about 30 lines down in your post.

8. Apr 15, 2009

### meiso

When a post is written in response to a user asking for homework help, I would assume that the original poster is going to read the response in its entirety (which, from his response, it seems he did), so I don't think about whether a certain item will be hard to "find" while I'm writing it.

9. Apr 15, 2009

### Staff: Mentor

IMO, it's better to answer the question right at the beginning, and follow it with any explanatory material.

10. Apr 16, 2009

thank you!