1.4 Inverse Functions

Big idea

Every function so far has been a one-way street: put $x$ in, get $f(x)$ out. This section asks when the street runs both ways. If you know the output, can you recover the input, and is the recovery itself a function? When the answer is yes, that recovery is called the inverse, and it is the tool behind solving equations of every kind you meet later.

The obstruction is repetition. If two different inputs share an output, then knowing the output does not identify the input, and no rule can undo the function. A function that never repeats an output is called one-to-one, and that is exactly the condition for an inverse to exist. Graphically it is the horizontal line test, the mirror image of the vertical line test that decided whether a graph was a function at all.

When a function does repeat, you are not stuck; you restrict the domain until it stops repeating. Squaring is not invertible on all real numbers, but it is on $x \ge 0$, and the inverse there is the square root. That move is the whole story behind the inverse trigonometric functions: sine repeats endlessly, so a single rising stretch is chosen as the official domain and everything else is discarded.

Inverses matter in calculus for two reasons that arrive soon. First, the logarithm is defined as the inverse of the exponential, so the next section rests on this one. Second, the derivative of an inverse can be read off the derivative of the original, because reflecting a graph across the line $y = x$ turns a slope into its reciprocal. Get the bookkeeping of domains and ranges right now and those results will read as consequences rather than new rules.

Decoder

A function $g$ is the inverse of a function $f$ when $g(f(x)) = x$ for every $x$ in the domain of $f$ and $f(g(y)) = y$ for every $y$ in the domain of $g$.

Two equations, not one, and both are needed. Each says that running the pair in one order leaves you where you started, and they are separate claims because the two compositions live on different sets. The first ranges over the domain of $f$; the second ranges over the range of $f$, which is the domain of $g$. That swap is the structural fact of the section: the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

The notation is a trap. In $f^{-1}(x)$ the $-1$ is not an exponent, and $f^{-1}(x)$ is not $1/f(x)$. The reciprocal of $f(x) = x + 2$ is $1/(x+2)$; the inverse is $x - 2$. The same superscript does mean a reciprocal on $\sin^2 x$, which is why inverse trigonometric functions are safer written as $\arcsin$, $\arccos$ and $\arctan$.

The classic mistake is producing a formula and stopping. A formula for an inverse is only half an answer; the other half is the domain you are allowed to feed it, which you get by asking what values $f$ actually attains. Skipping that step is how students end up claiming that the inverse of $x^2$ is $\pm\sqrt{x}$, which is not a function at all.

Definitions and results

One-to-one. A function is one-to-one when different inputs always give different outputs, that is, $f(a) = f(b)$ forces $a = b$. Only one-to-one functions have inverses.

The horizontal line test. A function is one-to-one exactly when no horizontal line meets its graph more than once. A function that is increasing on its whole domain, or decreasing on its whole domain, automatically passes.

The inverse function. If $f$ is one-to-one, its inverse $f^{-1}$ is the function that sends each output of $f$ back to the input that produced it: $f^{-1}(y) = x$ exactly when $f(x) = y$. Domain and range trade places.

Finding a formula. Write $y = f(x)$, solve the equation for $x$ in terms of $y$, then rename the variables so the answer reads as a function of $x$. Solving is the real work; renaming is cosmetic.

Graphs. The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$, because the reflection swaps the coordinates of every point. A point $(a,b)$ on one graph corresponds to $(b,a)$ on the other.

Restricting the domain. A function that fails the horizontal line test can be cut down to an interval on which it does not repeat, and the restricted function has an inverse. Different choices of interval give different inverses, so the choice must be stated.

Inverse trigonometric functions. With sine restricted to $[-\pi/2, \pi/2]$, cosine to $[0, \pi]$ and tangent to $(-\pi/2, \pi/2)$, each becomes one-to-one and the inverses are

$$ \arcsin: [-1,1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], \qquad \arccos: [-1,1] \to [0, \pi], \qquad \arctan: \mathbf{R} \to \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) $$

Cancellation, with fine print. $\sin(\arcsin x) = x$ holds for every $x$ in $[-1,1]$, but $\arcsin(\sin\theta) = \theta$ only when $\theta$ already lies in $[-\pi/2, \pi/2]$. Outside that window the composition returns the angle in the window with the same sine.

Worked examples

A formula from solving

Let $f(x) = 2x^3 - 5$. Cubing is increasing, so $f$ is increasing and passes the horizontal line test.

Set $y = 2x^3 - 5$ and solve for $x$. Add $5$, divide by $2$, take the cube root:

$$ x = \sqrt[3]{\frac{y + 5}{2}}, \qquad \text{so} \qquad f^{-1}(x) = \sqrt[3]{\frac{x + 5}{2}} $$

Cube roots accept every real number, and $f$ attains every real value, so both domain and range are all of $\mathbf{R}$ here and no restriction is needed.

Check: $f(1) = 2 - 5 = -3$, and $f^{-1}(-3) = \sqrt[3]{2/2} = 1$.

Verifying a proposed inverse

Claim: the inverse of $f(x) = \dfrac{2x - 1}{x + 3}$ is $g(x) = \dfrac{3x + 1}{2 - x}$.

Verify one composition in full. Substitute $g(x)$ into $f$ and clear the inner fractions by treating $2 - x$ as a common denominator:

$$ 2g(x) - 1 = \frac{6x + 2 - (2 - x)}{2 - x} = \frac{7x}{2 - x}, \qquad g(x) + 3 = \frac{3x + 1 + 3(2 - x)}{2 - x} = \frac{7}{2 - x} $$

Dividing these, the common denominator cancels and the sevens cancel, leaving $f(g(x)) = x$. The other composition works the same way. The domains confirm the picture: $f$ excludes $x = -3$ and never attains the value $2$, while $g$ excludes $x = 2$ and never attains $-3$.

Restricting to get an inverse

Let $f(x) = x^2 - 6x + 5$ with the domain restricted to $x \ge 3$.

Complete the square: $f(x) = (x - 3)^2 - 4$. The restriction starts at the vertex, so on $x \ge 3$ the function is increasing and one-to-one, and its outputs run over $[-4, \infty)$.

Solve $y = (x-3)^2 - 4$. Add $4$, take the square root, and choose the positive root because $x \ge 3$:

$$ x = 3 + \sqrt{y + 4}, \qquad \text{so} \qquad f^{-1}(x) = 3 + \sqrt{x + 4}, \quad x \ge -4 $$

Check: $f(5) = 25 - 30 + 5 = 0$, and $f^{-1}(0) = 3 + \sqrt{4} = 5$. Had the restriction been $x \le 3$ instead, the minus root would have been the correct choice and the inverse would be $3 - \sqrt{x+4}$.

Inverse trigonometric values

Evaluate $\arcsin(-1/2)$. You need the angle in $[-\pi/2, \pi/2]$ whose sine is $-1/2$, which is $-\pi/6$. Note that $7\pi/6$ also has sine $-1/2$ but is outside the official range, so it is not the answer.

Evaluate $\arccos(-\sqrt{2}/2)$. The range is $[0,\pi]$ and cosine is negative on the upper left of the circle, giving $3\pi/4$. Similarly $\arctan(\sqrt{3}) = \pi/3$.

Now a composition with fine print: $\arcsin\big(\sin(5\pi/6)\big)$. The inner value is $1/2$, and the angle in $[-\pi/2, \pi/2]$ with sine $1/2$ is $\pi/6$, not $5\pi/6$.

Finally, a mixed composition: $\cos(\arcsin(3/5))$. Call the inner angle $\theta$, so $\sin\theta = 3/5$ with $\theta$ in $[-\pi/2,\pi/2]$, where cosine is nonnegative. Then $\cos\theta = \sqrt{1 - 9/25} = 4/5$.

Practice

Start with existence and construction. Decide whether a function is one-to-one, then solve for the inverse and state its domain.

Next, confirmation. Compose the two functions in both orders and simplify until the input reappears.

Last, the restricted-domain case that matters most. Keep the official range of each inverse trigonometric function in view and let it choose the angle.

Quiz

Six items on one-to-one functions and inverse formulas, verification by composition, and values of the inverse trigonometric functions.