Inverses of Power Functions Are Power Functions

Power Function Inverses

The big takeaway for this section is exactly what the title says: inverse of power functions are power functions. There is a generalized process to show this, but it can be a little confusing so let's look at this with a simple example, $f(x)=x^2$. We're going to switch the variables and solve for $y$ the same way we would find any other inverse.

$$\begin{align}f(x)&=x^2\\ y&=x^2\\ x&=y^2\\ \sqrt{x}&=y\\ f^{-1}(x)&=\sqrt{x} \end{align}$$

Algebraically, we've shown that $f(x)=\sqrt{x}$ is the inverse of $f(x)=x^2$. This doesn't look like a power function in it's current form, but remember that roots are just functions to a power of a fraction. $f(x)=\sqrt{x}$ can be rewritten as $f(x)=x^{\frac{1}{2}}$, which means that square root functions are also power functions!

You may have learned before that even power functions like quadratics can't have inverses, but they actually can! We've seen here that their inverses are power functions, but there is a little more to it. The next step is to learn how restricting the domain can affect both the original and inverse function.

So how, exactly, do we find the inverse of a power function? You might remember our trick for finding inverses algebraically. But I'll recap now.

Say we have a function

$$y = f(x).$$

This is an EQUATION we can use to find the inverse $f^{-1}(x),$ if it exists (whether or not it does is another matter I'll get to later). We replace every appearance of $x$ with a $y$ and vice-versa, turning our equation into

$$x = f(y).$$

Now we solve this equation to get $y$ by itself on one side, and then the thing that remains on the other side of the equation is our inverse function (it needs to only be in terms of $x$ and not $y.$).

$$f^{-1}(x) = y$$

is our equation when all is said and done. It's just a matter of what this equation looks like for a power function.

Our most beloved example of a power function is

$$f(x) = x^3.$$

Let's execute the procedure we just talked about on this function. Set $y = f(x)$ and our equation becomes

$$y = x^3.$$

Now flip the $x$ and the $y:$

$$x = y^3.$$

Our goal is to solve for $y$ again. Luckily it's as simple as taking a cube root here:

$$y = \sqrt[3]{x}.$$

This means that $f^{-1}(x) = \sqrt[3]{x}.$ We can write this as $f^{-1}(x) = x^{\frac{1}{3}},$ which is another power function!

It so turns out that the inverse of any power function is also a power function. That's because the process of solving this equation follows the same general pattern for any power function. With that being said, let's do another example.

Example: Find the inverse of $f(x) = 4x^2.$

Let's do the same procedure again. Have $y = 4x^2,$ and then flip the $x$s and $y$s to get

$$x = 4y^2.$$

and solve for $y.$ Divide both sides by $4:$

$$\frac{x}{4} = y^2.$$

Now take the square root of both sides:

$$\pm \sqrt{\frac{x}{4}} = \sqrt{y^2} \Rightarrow$$ $$\pm \frac{1}{2}\sqrt{x} = y.$$

So which sign do we take, the $+$ or the $-?$ This is a problem whenever we deal with a power function that has an even exponent. To answer this question we have to consider the range of the original function. The original function was $y = x^2,$ which only has positive outputs. This means the inverse function must copy that behavior, so we use the $+.$ In other words,

$$\frac{1}{2}\sqrt{x} = y.$$

In conclusion, this time $f^{-1}(x) = \frac{1}{2}\sqrt{x}.$ But there's one other problem: this function is only defined when $x$ is nonnegative. That means we have to restrict the domain of the original function before we start! In other words, the question should have said:

Find $f^{-1}(x)$ if $f(x) = x^2$ on the domain $[0, \infty).$

With that, we're good to go. If the original power function has an odd exponent, this extra step is unnecessary.