Finding f^-1'(2): Solving a Homework Equation

[Joe_K]

Homework Statement

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

Homework Equations

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

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.

 

[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.

 

[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.

 

[SammyS]

Homework Statement

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

Homework Equations

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

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.

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".

 

[jackmell]

Homework Statement

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

Homework Equations

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

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.

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.