Finding the Inverse of a Cubic Polynomial

[QuarkCharmer]

Homework Statement

Find the inverse of the function.

[tex]f(x)=2x^{3}+5[/tex]

Homework Equations

Possibly the quadratic equation.

The Attempt at a Solution

[tex]f(x)=2x^{3}+5[/tex]

[tex]y=2x^{3}+5[/tex]

[tex]-2x^{3}=-y+5[/tex]

[tex]x^{3}= \frac{-y+5}{-2}[/tex]

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

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

and

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

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

 

[Mentallic]

You're confusing [tex]x^2=1 \to x=\pm 1[/tex] with [tex]x^3=1 \to x=1[/tex]

x=-1 does not satisfy [tex]x^3=1[/tex]

 

[QuarkCharmer]

Ah, yes it seems I am.

The inverse function is simply:

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

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

[tex]f(x)=2x^{3}+5[/tex]

is

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

 

[Mark44]

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

 

[QuarkCharmer]

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

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!