As we saw in the section on solving equations, the ability to “undo” functions is a key skill to have when solving equations.

Suppose you have to solve for $x$ in the equation \[ f(x) = c. \] where $f$ is some function and $c$ is some constant. Our goal is to isolate $x$ on one side of the equation but there is the function $f$ standing in our way.

The way to get rid of $f$ is to apply the inverse function (denoted $f^{-1}$) which will “undo” the effects of $f$. We find that: \[ f^{-1}\!\left( f(x) \right) = x = f^{-1}\left( c \right). \] By definition the inverse function $f^{-1}$ does the opposite of what the function $f$ does so together they cancel each other out. We have $f^{-1}(f(x))=x$ for any number $x$.

Provided everything is kosher (the function $f^{-1}$ must be defined for the input $c$), the manipulation we made above was valid and we have obtained the answer $x=f^{-1}( c)$.

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Note the new notation for denoting the function inverse $f^{-1}$ that we introduced in the above example. This notation is borrowed from the notion of “inverse number”. Multiplication by the number $d^{-1}$ is the inverse operation of multiplication by the number $d$: $d^{-1}dx=1x=x$. In the case of functions, however, the negative one exponent does not mean the inverse number $\frac{1}{f(x)}=(f(x))^{-1}$ but functions inverse, i.e., the number $f^{-1}(y)$ is equal to the number $x$ such that $f(x)=y$.

You have to be careful because sometimes the applying the inverse leads to multiple solutions. For example, the function $f(x)=x^2$ maps two input values ($x$ and $-x$) to the same output value $x^2=f(x)=f(-x)$. The inverse function of $f(x)=x^2$ is $f^{-1}(x)=\sqrt{x}$, but both $x=+\sqrt{c}$ and $x=-\sqrt{c}$ would be solutions to the equation $x^2=c$. A shorthand notation to indicate the solutions for this equation is $x=\pm c$.

Here is a list of common functions and their inverses:

\[ \begin{align*} \textrm{function } f(x) & \ \Leftrightarrow \ \ \textrm{inverse } f^{-1}(x) \nl x+2 & \ \Leftrightarrow \ \ x-2 \nl 2x & \ \Leftrightarrow \ \ \frac{1}{2}x \nl -x & \ \Leftrightarrow \ \ -x \nl x^2 & \ \Leftrightarrow \ \ \pm\sqrt{x} \nl 2^x & \ \Leftrightarrow \ \ \log_{2}(x) \nl 3x+5 & \ \Leftrightarrow \ \ \frac{1}{3}(x-5) \nl a^x & \ \Leftrightarrow \ \ \log_a(x) \nl \exp(x)=e^x & \ \Leftrightarrow \ \ \ln(x)=\log_e(x) \nl \sin(x) & \ \Leftrightarrow \ \ \arcsin(x)=\sin^{-1}(x) \nl \cos(x) & \ \Leftrightarrow \ \ \arccos(x)=\cos^{-1}(x) \end{align*} \]

The function-inverse relationship is reflexive. This means that if you see a function on one side of the above table (no matter which), then its inverse is on the opposite side.

Let's say your teacher doesn't like you and right away on the first day of classes, he gives you a serious equation and wants you to find $x$: \[ \log_5\left(3 + \sqrt{6\sqrt{x}-7} \right) = 34+\sin(5.5)-\Psi(1). \] Do you see now what I meant when I said that the teacher doesn't like you?

First note that it doesn't matter what $\Psi$ is, since $x$ is on the other side of the equation. We can just keep copying $\Psi(1)$ from line to line and throw the ball back to the teacher in the end: “My answer is in terms of your variables dude. You have to figure out what the hell $\Psi$ is since you brought it up in the first place.” The same goes with $\sin(5.5)$. If you don't have a calculator, don't worry about it. We will just keep the expression $\sin(5.5)$ instead of trying to find its numerical value. In general, you should try to work with variables as much as possible and leave the numerical computations for the last step.

OK, enough beating about the bush. Let's just find $x$ and get it over with! On the right side of the equation, we have the sum of a bunch of terms and no $x$ in them so we will just leave them as they are. On the left-hand side, the outer most function is a logarithm base $5$. Cool. No problem. Looking in the table of inverse functions we find that the exponential function is the inverse of the logarithm: $a^x \Leftrightarrow \log_a(x)$. To get rid of the $\log_5$ we must apply the exponential function base five to both sides: \[ 5^{ \log_5\left(3 + \sqrt{6\sqrt{x}-7} \right) } = 5^{ 34+\sin(5.5)-\Psi(1) }, \] which simplifies to: \[ 3 + \sqrt{6\sqrt{x}-7} = 5^{ 34+\sin(5.5)-\Psi(1) }, \] since $5^x$ canceled the $\log_5 x$.

From here on it is going to be like if Bruce Lee walked into a place with lots of bad guys. Addition of $3$ is undone by subtracting $3$ on both sides: \[ \sqrt{6\sqrt{x}-7} = 5^{ 34+\sin(5.5)-\Psi(1) } - 3. \] To undo a square root you take the square \[ 6\sqrt{x}-7 = \left(5^{ 34+\sin(5.5)-\Psi(1) } - 3\right)^2. \] Add $7$ to both sides \[ 6\sqrt{x} = \left(5^{ 34+\sin(5.5)-\Psi(1) } - 3\right)^2+7. \] Divide by $6$: \[ \sqrt{x} = \frac{1}{6}\left(\left(5^{ 34+\sin(5.5)-\Psi(1) } - 3\right)^2+7\right), \] and then we square again to get the final answer: \[ \begin{align*} x &= \left[\frac{1}{6}\left(\left(5^{ 34+\sin(5.5)-\Psi(1) } - 3\right)^2+7\right) \right]^2. \end{align*} \]

Did you see what I was doing in each step? Next time a function stands in your way, hit it with its inverse, so that it knows not to ever challenge you again.

The recipe I have outlined above is not universal. Sometimes $x$ isn't alone on one side. Sometimes $x$ appears in several places in the same equation so can't just work your way towards $x$ as shown above. You need other techniques for solving equations like that.

The bad news is that there is no general formula for solving complicated equations. The good news is that the above technique of “digging towards $x$” is sufficient for 80% of what you are going to be doing. You can get another 15% if you learn how to solve the quadratic equation: \[ ax^2 +bx + c = 0. \]

Solving third order equations $ax^3+bx^2+cx+d=0$ with pen and paper is also possible, but at this point you really might as well start using a computer to solve for the unknown(s).

There are all kinds of other equations which you can learn how to solve: equations with multiple variables, equations with logarithms, equations with exponentials, and equations with trigonometric functions. The principle of digging towards the unknown and applying the function inverse is very important so be sure to practice it.