In case that doesn’t show up:

The function $f(x) = x + \sqrt{x}$ is one-to-one. Find its inverse.

That’s a nice challenge. My approach – which is clearly a spoiler – is below the line.

I tend to start function-inverses by renaming my variables for convenience. The graph of the inverse function is a reflection in $y=x$, so I usually rename $x$ to $y$, $f(x)$ to $x$, and try to rearrange.

So I have $x = y + \sqrt{y}$, and I need to find an expression for $y$ in terms of $x$.

I spy a hidden quadratic. If I let $z = \sqrt{y}$, then I have $z^2 + z - x = 0$.

I could throw the equation at it, but that’s not how we roll around here.

• $4z^2 + 4z - 4x = 0$
• $(2z+1)^2 = 4x + 1$
• $2z+1 = \pm \sqrt{4x+1}$
• $2z = -1 \pm \sqrt{4x+1}$
• $4z^2 = 1 \pm 2\sqrt{4x+1} + (4x+1)$
• $4y = 2 \pm 2\sqrt{4x+1} + 4x$

So the inverse relation is $y = x + \frac{1}{2}\left(1 \pm \sqrt{4x+1}\right)$.

But that’s not the inverse function. It’s not a function at all – it’s double-valued.

Because $f(0)=0$, we need our final function to go through $(0,0)$ as well – and the negative branch is the one that works, so our function is:

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

Only we’re not quite done. As it stands, $f^{-1}\left(-\frac{1}{4}\right)= \frac{1}{4}$, and $f\left(\frac{1}{4}\right)$ certainly isn’t $-\frac{1}{4}$! ((I think this is an artifact of all the squaring and square-rooting in the process.))

We need to restrict the domain to $x \ge 0$ to make sure the inverse function is also one-to-one.

I like that there were a couple of wrinkles to this one – and there may be more I didn’t spot! Did you tackle it a different way? Did you get a different solution? As always, I love to hear about these things!