Finding the inverse of $f(x)=x(x+1)$.

Question

This blog entry Inverse functions: we're teaching it all wrong, warns against swapping variables when finding the inverse of a function.

What would be the general procedure for finding the inverse of $f(x)=x(x+1)$ (domain: $\mathbb{R}$, without swapping dependent and independent variables ?

Answer 1

I will not comment on this "variable swapping" controversy.

Write

$$y=x(x+1)=\left(x+\frac12\right)^2-\frac14$$ yields the solution

$$x=-\frac12\pm\sqrt{y+\frac14}.$$

From this it is clear that we should have $y\ge -\dfrac14$ and that there are two distinct solutions in $y$. Choosing one of the branches, we have the inverse function

$$f^{-1}(x)=-\frac12+\sqrt{x+\frac14}, x\ge-\dfrac14$$

where

$$f(x)=x(x+1), x\ge-\frac12.$$

Answer 2

You will first have to define a domain to make the function injective, before finding its inverse.

eg. If you take the domain of $f(x)$ as $\left[ \frac{-1}{2}, \infty \right)$

$$y=x(x+1)$$

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

$$y=(x+\frac{1}{2})^2-\frac{1}{4}$$

$$x=-\frac{1}{2} \pm \sqrt{y+\frac{1}{4}}$$

Now, since the domain is restricted to be $> \frac{-1}{2}$ we are down to one branch of the $\pm$, specifically the $+$ one.

So,

$$f^{-1}(x)=-\frac{1}{2} +\sqrt{x+\frac{1}{4}}$$

Similarly, if the domain was $(-\infty,-\frac{1}{2}]$, the $-$ branch would be selected.

Here, $$f^{-1}(x)=-\frac{1}{2}-\sqrt{x+\frac{1}{4}}$$ $${}$$