Composition and inverse functions

A coffee machine works in two stages: it grinds the beans, then brews them. Put beans in and a drink comes out — but the order cannot be swapped, and no cup gives the beans back. Composition and the inverse function are exactly those two questions.

Composing functions

The output of one function can be the argument of another:

The notation is read from the inside out: $g$ acts first, then $f$. We call $g$ the inner function and $f$ the outer one.

The formula of a composition

Instead of computing point by point, substitute the whole inner function for the argument:

Two different polynomials — formal confirmation that composition is not commutative:

The chain rule in the lesson on the derivative differentiates expressions of exactly this shape — and only now is it clear what they are.

The domain of a composition

This is where mistakes are easiest. The domain of $f \circ g$ consists of those arguments $x$ for which:

1. $x$ lies in the domain of $g$, and
2. the value $g(x)$ lies in the domain of $f$.

The second condition can narrow things sharply. For $f(x) = \sqrt{x}$ and $g(x) = x - 3$ the function $g$ is defined everywhere, but the composition $\sqrt{x-3}$ needs $x - 3 \ge 0$, so the domain is $[3,\, \infty)$.

Monotonicity of a composition

A composition inherits its monotonicity from its parts — and the proof is short, because it fits entirely inside the definition.

Theorem. If $g$ is increasing and $f$ is increasing, then $f \circ g$ is increasing.

Proof. Take any $x_1 < x_2$ in the domain of the composition.

The inequality between the arguments passed through both functions untouched, so $(f \circ g)(x_1) < (f \circ g)(x_2)$. That is exactly the definition of an increasing function. $\blacksquare$

The same scheme with the inequalities reversed gives the remaining three cases:

The last row surprises, but it behaves like multiplying negative numbers: two reversals cancel each other out.

One-to-one functions

For a function to be undoable, every value must come from exactly one argument:

Such a function is called one-to-one. On a graph this is checked by the horizontal-line test: if any horizontal line meets the graph more than once, the function is not one-to-one.

The function $y = x^2$ on the whole line fails the test — the line $y = 4$ hits it twice. Every strictly monotone function passes it, because a direction once chosen is never reversed.

The inverse function

For a one-to-one function there is a function $f^{-1}$ that undoes $f$:

Domain and range swap roles in the process: what used to be the output becomes the input.

Mind the notation. $f^{-1}$ does not mean $\tfrac{1}{f}$. For $f(x) = 2x$ we have $f^{-1}(x) = \tfrac{x}{2}$, while $\tfrac{1}{f(x)} = \tfrac{1}{2x}$ — two entirely different functions.

Finding the formula

Three steps, always the same:

1. write $y = f(x)$;
2. solve that equation for $x$;
3. swap the names of the variables.

The graph

Since $f$ maps $a \mapsto b$, the inverse maps $b \mapsto a$. The point $(a,\, b)$ therefore goes to $(b,\, a)$, and swapping the coordinates is a reflection in the line $y = x$.

When the domain has to be restricted

A function that fails the horizontal-line test is not lost — it is enough to cut part of it away.

The same move stands behind three other pairs you have met or will meet:

• $y = x^2$ for $x \ge 0$ and the square root $y = \sqrt{x}$;
• $y = b^x$ and the logarithm $y = \log_b x$ — here no restriction is needed, because the exponential function is one-to-one on the whole line;
• $y = \sin x$ restricted to $\left[-\tfrac{\pi}{2},\, \tfrac{\pi}{2}\right]$ and the inverse trigonometric functions, that is arcsine and its siblings.

The last case is the best test of this lesson: the sine repeats every $2\pi$, so the equation $\sin x = \tfrac{1}{2}$ has infinitely many solutions. Only restricting the domain to a single full run makes the function one-to-one and therefore invertible — and the graph of $\arcsin$ is the reflection of that restricted sine in the line $y = x$, exactly as in the picture above.

Practice

The set checks four skills: the value of a composition $f(g(x))$ at a point (work from the inside out), the formula of a composition after substitution, the formula of an inverse function, and one value of it — the last of which needs no formula at all, only the question "for which $x$ does the function give this value". Formulas may be typed in any equivalent spelling: (x−3)/2 and 0.5x−1.5 are both accepted. Proving the monotonicity of a composition stays with the theorem above — its answer is a chain of reasoning, not a number.

Common mistakes

• Reading $f(g(x))$ from the left — always start with the inner function, that is with $g$.
• Assuming commutativity — $f \circ g$ and $g \circ f$ are in general two different functions.
• Skipping the domain of the composition — "$g$ is defined everywhere" is not enough; the value $g(x)$ has to land in the domain of $f$.
• Treating $f^{-1}$ as $\tfrac{1}{f}$ — it is the inverse function, not the reciprocal.
• Inverting a function that is not one-to-one — without restricting the domain there is no inverse.
• Forgetting the third step — after solving for $x$ the variables have to be renamed so that the argument is called $x$ again.