Simply put, inverse functions are functions with the $x$ and $y$ reversed, and they are often denoted by $f^{-1}$. To find an inverse function, switch the variables are rearrange. So, if $f(x) = 2x + 3$ then we do

\[\begin{align*} x &= 2y + 3 \\ y &= \frac{1}{2}x - \frac{3}{2} \\ f^{-1}(x) &= \frac{1}{2}x - \frac{3}{2} \end{align*}\]And you can verify that the two functions are inverses of each other using function composition. If they are inverses, then $f(g(x)) = x$ and also $g(f(x)) = x$. Checking one is enough to prove inversality (not a real word).

\[\begin{align*} f(g(x)) &= 2\left(\frac{1}{2}x - \frac{3}{2}\right) + 3 \\ &= x - 3 + 3 \\ &= x \end{align*}\]You can think of inverse functions as undoing the original, so you are left with the just $x$. Related to this, when graphed you’ll notice that the inverse function is the original reflected over the line $y=x$. You can see our pair graphed on Desmos here.

Everything above is the simple version, whereas reality is more complicated. When you invert a function, it’s domain and range are swapped and any restrictions go along with it. The function $x^2$ does not have an inverse function, as its actual inverse is not a function.

So, a function has to be one-to-one in order for it to have an inverse, meaning each $x$ value maps to exactly one $y$ value (which is the definition of a function), but also each $y$ value matches to exactly one $x$ value. You can use the Horizontal Line Test to see if a function has an inverse. It works exactly like the Vertical Line Test, except horizontal.

Technically, trigonometric functions can’t have inverse functions. The oscillate and fail the Horizontal Line Test. But, inverse trigonometric functions do exist. You can solve $\sin x = 1$ by applying $\arcsin$ (or $\sin^{-1}$) to both sides.

The inverse trig functions are actually built off of trig functions with restricted domains. Since $\sin$ repeats, we can just use one period of the function to generate the inverse. Instead of using $\sin$ with its normal domain of all real numbers, we restrict it to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, giving us $\arcsin$ with a domain of $[-1,1]$ and a range of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

This does lead to extra steps if you need to apply an inverse trig function involving an angle outside of its range, since now you have to find an equivalent angle. Luckily, its just adding or subtracting $2\pi$ (360º) until we get back in range … and also this is isn’t a trig class so it won’t come up often.