# Inverse Functions: Solving Equations (Part 4)

Although disguised, inverse functions play an important role in the ordinary solution of equations. For example, consider the steps used to solve this simple algebra problem:

$2x + 4 = 10$

$2x = 6$

$x = 3$

To go from the first equation to the second equation, let $X_1 = 2x+4$ and $X_2 = 10$, and let $f(x) = x – 4$. This is an bijective function with inverse $f^{-1}(x) = x +4$. Therefore,

$X_1 = X_2 \quad \Longleftrightarrow \quad f(X_1) = f(X_2)$

Stated another way,

$2x + 4 = 10 \quad \Longleftrightarrow 2x = 6$

Again, let $X_3 = 2x$ and $X_4 = 6$, and let $g(x) = x/2$. This is also a bijective function with inverse function $g^{-1}(x) = 2x$. Therefore,

$X_3= X_4 \quad \Longleftrightarrow \quad g(X_1) = g(X_2)$

Stated another way,

$latex 2x + 4 = 10 \quad \Longleftrightarrow 2x = 6 \quad \Longleftrightarrow x = 3$

So we are guaranteed that $x= 3$ is the one and only one solution of this equation.

If the process of solving an equation requires the use of a function that isn’t a bijection, then funny things can happen. For example, consider the slightly more complicated equation

$\sqrt{x} = x - 6$

Let’s starting solving by squaring both sides:

$x = (x-6)^2$

$x = x^2 - 12x + 36$

$0 = x^2 - 13x + 36$

$0 = (x-9)(x-4)$

$x - 9 = 0 \quad \hbox{or} \quad x - 4 = 0$

$x = 9 \quad \hbox{or} \quad x = 4$

So there are two solutions, right? Well…

$\sqrt{9} = 3 = 9 - 6$,

but $\sqrt{4} \ne 4 - 6$!

So what happened? In other words, what is qualitatively different about this problem that didn’t happen in the first problem to produce an extraneous solution? The problem is the first step. Let $X_1 = \sqrt{x}$ and $X_2 = x-6$. We applied the function $f(x) = x^2$ to both sides. Unfortuntely, $f(x) = x^2$ is not an invertible function when using the entire real line as the domain of $f$. In other words,

$\sqrt{x} = x -6 \quad$ implies $\quad x = (x-6)^2$,

but $x =(x-6)^2 \quad$ does not imply that $\quad \sqrt{x} = x - 6$.

The practical upshot is that, when arriving at the final step of the solution, we can’t be certain that the “solutions” we obtain actually work. Instead, what we’ve really shown that anything other than the solutions can’t work, which is different than saying that these two solutions actually do work. So it remains to actually check that these potential solutions are actually solutions (or not).

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