# Thread: Verifying a function has an inverse

1. ## Verifying a function has an inverse

Verify $f(x) = 1 - 2x - x^3$ has an inverse, and find $(f^{-1})'(4)$.

So I take the derivative and get $f'(x) = -2 -3x^2$.

It's negative for all x, so it's always decreasing, so it's 1-to-1 and has an inverse.

Now, how do I find $(f^{-1})'(4)$? Is this the same as $\frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50}$?

2. ## Re: Verifying a function has an inverse

To find the inverse function $f^{-1}$. If $y=1-2x-x^3$ then replace the $x$ and $y$ so $x=1-2y-y^3$, now find an expression $y=...$ in function of $x$. So you will get the inverse function.

3. ## Re: Verifying a function has an inverse

Originally Posted by deezy
Verify $f(x) = 1 - 2x - x^3$ has an inverse, and find $(f^{-1})'(4)$.

So I take the derivative and get $f'(x) = -2 -3x^2$.

It's negative for all x, so it's always decreasing, so it's 1-to-1 and has an inverse.

Now, how do I find $(f^{-1})'(4)$? Is this the same as $\frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50}$?

For the inverse part:

Prove that f is one-to-one an onto.

4. ## Re: Verifying a function has an inverse

Originally Posted by Also sprach Zarathustra
For the inverse part:

Prove that f is one-to-one an onto.
And also look here: Inverse functions and differentiation - Wikipedia, the free encyclopedia

5. ## Re: Verifying a function has an inverse

Originally Posted by deezy
Verify $f(x) = 1 - 2x - x^3$ has an inverse, and find $(f^{-1})'(4)$.
Because $f(-1)=4$ we need

$(f^{-1})'(4)=\frac{1}{f'(-1)}$.

6. ## Re: Verifying a function has an inverse

How do you determine that $f(-1)=4$?

Switching the x and y wasn't helping because there's 2 ys in the equation.
$x=1-2y-y^3$
$-2y-y^3 = x - 1$???

7. ## Re: Verifying a function has an inverse

Originally Posted by deezy
How do you determine that $f(-1)=4$?
Switching the x and y wasn't helping because there's 2 ys in the equation.
$x=1-2y-y^3$
$-2y-y^3 = x - 1$???
I do not think you can solve that inverse.
You just find $f(x)=4$ for $x$.

8. ## Re: Verifying a function has an inverse

So
$1-2x-x^3=4$
$x^3 + 2x = 3$
$x = -1$.

I can see that instinctively... but if it's a harder equation I will just have to use trial and error?

9. ## Re: Verifying a function has an inverse

Originally Posted by deezy
So
$1-2x-x^3=4$
$x^3 + 2x = 3$
$x = -1$.

I can see that instinctively... but if it's a harder equation I will just have to use trial and error?
Well that is just the way it is. Often a graph will give you an idea of the pre-image

10. ## Re: Verifying a function has an inverse

@ Deezy:

Sorry, I just wanted to give a 'general' solution to calculate the inverse function, but indeed in this case that would be really hard because of the third degree, if you have a first degree or second degree function it's less complicated.

11. ## Re: Verifying a function has an inverse

In this case it's not very hard. We have the equation $x^3+2x+y-1=0$ i.e. $p=2,q=y-1$ . Then, $D=-108\left(\frac{q^2}{4}+\frac{p^3}{27}\right)<0$ which implies the equation has only one real root. So, using the well known formula for the roots:

$x=\sqrt[3]{-\frac{y-1}{2}+\sqrt{\frac{(y-1)^2}{4}+\frac{8}{27}}}+\sqrt[3]{-\frac{y-1}{2}-\sqrt{\frac{(y-1)^2}{4}+\frac{8}{27}}}$

We can inmediately verify that the product of the two real cubic roots of the above addends is $-2/3=-p/3$ . So, we can express

$f^{-1}(x)=\sqrt[3]{-\frac{x-1}{2}+\sqrt{\frac{(x-1)^2}{4}+\frac{8}{27}}}+\sqrt[3]{-\frac{x-1}{2}-\sqrt{\frac{(x-1)^2}{4}+\frac{8}{27}}}$

where we choose for each addend the real cubic root.

12. ## Re: Verifying a function has an inverse

Originally Posted by Also sprach Zarathustra
For the inverse part:

Prove that f is one-to-one an onto.

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

$f'(x)=-2-3x^2$

$f'(x)<0$ for all $x\in\mathbb{R}$

In other words $f(x)$ is monotone decreasing for all $x\in\mathbb{R}$.

We deduce from the above that $f(x)$ is bijection.

To show that $f(x)$ is surjective, we note that:

$\lim_{x\to\infty}f(x)=-\infty$

and,

$\lim_{x\to-\infty}f(x)=\infty$

These two results with MVT(with a little work...) we wil get that $f(x)$ is surjective function.