1. ## Inverse functions

How do I find f^-1 (2), x for f^-1 (x) = 0

I have a graph with a function f on it,

f(x) = [ 2x^3 + 1/2 x + 5 ]/2

And also, it would be great if someone could find the equation for the inverse of f(x)

2. Originally Posted by realintegerz
How do I find f^-1 (2), x for f^-1 (x) = 0

I have a graph with a function f on it,

f(x) = [ 2x^3 + 1/2 x + 5 ]/2

And also, it would be great if someone could find the equation for the inverse of f(x)
Remember that for inverses the domain and range switch. For example

$\displaystyle f(x)$ has domain A and range B than $\displaystyle f^{-1}(x)$ has domain B and range A

Ok so using this info we can solve what $\displaystyle f^{-1}(x) = 0$ and $\displaystyle f^{-1}(2)$ is since we know the function is

$\displaystyle f(x) = \frac{2x^3 +\frac{x}{2} +5}{2}$

1) $\displaystyle f^{-1}(x) = 0$ is $\displaystyle f(0)=?$

so

$\displaystyle x = \frac{5}{2}$

2)To solve the other problem set

$\displaystyle 2 = \frac{2x^3 +\frac{x}{2} +5}{2}$

and solve for x

3. hmmm..

im stuck on that f^-1 (2)...

So far i simplified it to...

-2 = 4x^3 + x

4. Originally Posted by realintegerz
hmmm..

im stuck on that f^-1 (2)...

So far i simplified it to...

-2 = 4x^3 + x
i tried using the rational root theorem but couldn't get any nice roots so i used the cubic equation. You could also do it numerically.

5. Originally Posted by 11rdc11
i tried using the rational root theorem but couldn't get any nice roots so i used the cubic equation. You could also do it numerically.
I know the quadratic equation..but not cubic...

could you explain that?