Evaluating the inverse of a function

[Umar]

If f(x) = x^5 + x^3 + x, find f^-1 (3).

I know we have to set the function equal to 3, and solve for x, but I don't think you can simplify the right side. Any help?

 

[MarkFL]

I think, if given:

$$f(x)=x^5+x^3+x$$

I would look at:

$$f'(x)=5x^4+3x^2+1=5\left(x^4+\frac{3}{5}x^2\right)+1=5\left(x^4+\frac{3}{5}x^2+\frac{9}{100}\right)+1-\frac{9}{20}=5\left(x^2+\frac{3}{10}\right)^2+\frac{11}{20}$$

We see then that for all $x$, we have $f'>0$ and so $f$ is monotonically increasing, and thus $f(x)+C$ will have only 1 real root. And so:

$$f(x)=3$$

will have only 1 real-valued solution. By inspection we see that:

$$f(1)=3$$

Hence:

$$f^{-1}\left(f(1)\right)=f^{-1}\left(3\right)$$

or:

$$f^{-1}\left(3\right)=1$$

 

[HallsofIvy]

In general, it is very difficult to find an inverse function but, as MarkFL pointed out, for this particular function, it is easy to see that $$f(1)= 1^5+ 1^3+ 1= 3$$. The "hard part" was showing that this function has and inverse but, from your first post, it seems you were not required to do that.