1. ## Inverse function

f(x)=x^3 + 2x -1

How do I find the inverse function?

2. Originally Posted by Stuck Man
f(x)=x^3 + 2x -1

How do I find the inverse function?
not the easiest thing to find ... is there a specific reason for finding the inverse? in other words, is this need for an inverse function tied to another question?

3. It ask s for the value of the inverse function at 2. It then asks for the value of the derivative of the inverse function at 2.

4. Originally Posted by Stuck Man
It ask s for the value of the inverse function at 2. It then asks for the value of the derivative of the inverse function at 2.
that's what I thought ... you do not need the inverse function to do this problem.

$f(x) = x^3 + 2x - 1$

note that $f(1) = 2$ , so if $g(x)$ is the inverse of $f(x)$, then $g(2) = 1$

further, if two functions, $f(x)$ and $g(x)$, are inverses, then $f[g(x)] = x$

$\frac{d}{dx} \left(f[g(x)] = x\right)
$

$f'[g(x)] \cdot g'(x) = 1$

$g'(x) = \frac{1}{f'[g(x)]}
$

$g'(2) = \frac{1}{f'[g(2)]}$

$g'(2) = \frac{1}{f'[g(2)]}$

$g'(2) = \frac{1}{f'(1)}$

find $f'(1)$ and finish it.

5. I've not done any calculus like that so I don't understand it but I might someday.

6. What do you mean by "any calculus like that"? If you are asked this question, surely you have worked with derivatives and the chain rule? If g is the inverse function to f, then f(g(x))= x so, by the chain rule, f'(g(x))g'(x)= 1. g(2)= 1 because $f(x)= x^3+ 2x- 1= 2$ has 1 as an obvious root. f'(g(2))= f'(1) which is $3(1)^2+ 2= 5$. So 5g'(2)= 1, g'(2)= 1/5.