Finding the Inverse of a Cubic Polynomial

[QuarkCharmer]

Homework Statement

Find the inverse of the function.

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

Homework Equations

Possibly the quadratic equation.

The Attempt at a Solution

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

$$y=2x^{3}+5$$

$$-2x^{3}=-y+5$$

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

$$x= \pm\sqrt[3]{\frac{-y+5}{-2}}$$

$$y= \pm\sqrt[3]{\frac{-x+5}{-2}}$$So the solution is two inverse functions? like..$$f^{-1}(x)= \sqrt[3]{\frac{(-x+5)}{-2}}$$

and

$$f^{-1}(x)= -\sqrt[3]{\frac{(-x+5)}{-2}}$$

I'm not sure that is what the professor is looking for? Thank you.

 

[Mentallic]

You're confusing $$x^2=1 \to x=\pm 1$$ with $$x^3=1 \to x=1$$

x=-1 does not satisfy $$x^3=1$$

 

[QuarkCharmer]

Ah, yes it seems I am.

The inverse function is simply:

$$f^{-1}(x)= \sqrt[3]{\frac{(-x+5)}{-2}}$$

then?

 

[aim1732]

Since the function is monotonic increasing function it has only one inverse and you have it.

 

[QuarkCharmer]

So the correct inverse function for

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

is

$$f^{-1}(x)= \sqrt[3]{\frac{(-x+5)}{-2}}$$
?
Thanks!

 

[Mark44]

Yes, but that's the same as
$$f^{-1}(x)= \sqrt[3]{\frac{x-5}{2}}$$

 

[QuarkCharmer]

Yes, but that's the same as
$$f^{-1}(x)= \sqrt[3]{\frac{x-5}{2}}$$

Right!

Thanks a lot for the help. This just sort of showed up on a worksheet, and we have not covered inverse functions yet, I had to read ahead in the book to even get the slightest idea.

It's very much appreciated!