# Finding (f^-1)'a

1. Sep 11, 2011

### Joe_K

1. The problem statement, all variables and given/known data

f(x)= x^3 + 3sinx + 2cosx, a=2

2. Relevant equations

(f^-1)'(a)= 1 / f'(f^-1(a))

3. The attempt at a solution

I am lost as to where to start on this problem. Can someone get me going in the right direction? Thanks in advance.

2. Sep 11, 2011

### Dick

A good place to start would be to try to find f^(-1)(2). Your expression for f is hard to invert in general but you can find f^(-1)(2) by guessing.

3. Sep 11, 2011

### lineintegral1

It is not possible to solve for the inverse analytically (as far as I know at least), so think about what value of $x$ would yield $f(x)=2$. In other words, compute $f^{-1}(2)$ by inspection. This is not as hard as it seems because you've been given some trig functions that are easy to deal with for certain values of $x$. Then, it is trivial as it is simply plug and chug.

4. Sep 12, 2011

### SammyS

Staff Emeritus
What does your Relevant Equation, (f -1)'(a)= 1 / f '(f -1(a)) tell you to do ??

Finding f -1(a) is like solving your previous problem.

Then find f '(x), and as lineintegral1 said, "plug & chug".

5. Sep 12, 2011

### jackmell

That notation is just making it confusing and harder to deal with. May I suggest looking at it more intuitively to make it easier to deal with.

Let:

$$z=x^3+3\sin(x)+2\cos(x)$$

and therefore:

$$z_0=x_0^3+3\sin(x_0)+2\cos(x_0)$$

and therefore we can write:

$$\frac{dx}{dz}\biggr|_{z=z_0}=\displaystyle\frac{1}{\frac{dz}{dx}\biggr|_{x=x_0}}$$

Ok, so now at that point, just calculate a few that you already know just to get the feel of it. Say (approx):

$$4.605=(1)^3+3\sin(1)+2\cos(1)$$

so that I can write:

$$\frac{dx}{dz}\biggr|_{z=4.605}=\displaystyle\frac{1}{\frac{dz}{dx}\biggr|_{x=1}}=\frac{1}{3x^2+3\cos(x)-2\sin(x)}\biggr|_{x=1}=0.340$$

Now do a few more say let x=-2,-1,0,2 and see what happens.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook