Types of functions, inverses and composites

Domain, codomain, range and types of function

A function $f: A \to B$ is a rule that sends each element of the domain $A$ to exactly one element of the codomain $B$.

For $f(x) = x^2$ with $f:\mathbb{R}\to\mathbb{R}$, the domain and codomain are both $\mathbb{R}$, but the range is only $[0, \infty)$, because a square is never negative.

Three words describe how a function behaves:

The horizontal line test makes this visual. Draw horizontal lines across the graph:

• if no horizontal line cuts the graph more than once, $f$ is injective;

• if every horizontal line through the codomain cuts the graph at least once, $f$ is surjective.

The line $y = 2$ cuts the parabola at $x = \pm\sqrt{2}$. Two different inputs give the same output, so $f(x) = x^2$ on $\mathbb{R}$ is not injective. The line $y = -1$ never cuts it, so $f$ is not surjective onto $\mathbb{R}$.

By contrast, $f(x) = x^3$ on $\mathbb{R}\to\mathbb{R}$ is bijective: every horizontal line cuts the graph exactly once.

Completing the square

To find the range of a quadratic, or to make it injective by restricting its domain, you need to know where its turning point is. Completing the square gives it straight away.

The reason it works: $(x + \tfrac b2)^2$ expands to $x^2 + bx + \tfrac{b^2}{4}$, which has an extra $\tfrac{b^2}{4}$ that you must take away again.

Move the sliders to see how $(x-a)^2 + b$ has its turning point at $(a, b)$.

Composite functions

Hiring a boat, then applying a discount; converting °C to °F, then rounding: often one function is applied straight after another. This is a composite function.

The order matters. Work from the inside out.

For $f(g(x))$ to exist, every output of $g$ must be an allowed input of $f$. For example, with $f(x) = \sqrt{x}$ and $g(x) = x - 4$, $f(g(x)) = \sqrt{x-4}$ only makes sense for $x \ge 4$.

Inverse functions

The inverse $f^{-1}$ undoes $f$: if $f$ sends $a$ to $b$, then $f^{-1}$ sends $b$ back to $a$.

Why does it need to be bijective? If two inputs gave the same output, the inverse would not know which one to go back to. If some codomain value is never hit, the inverse would have nothing to send it to.

Because the graph of $f^{-1}$ is the reflection of the graph of $f$ in $y = x$, any point where the graph of $f$ meets the line $y = x$ is also on the graph of $f^{-1}$. So the graphs of $f$ and $f^{-1}$ meet the line $y = x$ at the same points, namely where $f(x) = x$.

The graphs below show $f(x) = 2x - 3$ (its inverse is $\frac{x+3}{2}$) and the mirror line $y = x$.

Some inverses need more algebra. Here the unknown appears twice.

Restricting the domain to get an inverse

A quadratic is not injective on $\mathbb{R}$, so it has no inverse there. But if you keep only one side of the turning point, it becomes bijective, and then an inverse exists. This is called restricting the domain (and the codomain, to the range).

To show a function is injective without a graph, assume $f(a) = f(b)$ and prove $a = b$. For the rational function above this works by cross-multiplying. For a quadratic on a restricted domain, completing the square shows which side of the turning point you are on.

Exponential and logarithmic functions as inverses

The functions $e^x$ and $\ln x$ are the standard example of an inverse pair.

The domain and range swap, as the inverse rule says. Their graphs are mirror images in $y = x$.

To find the inverse of a function involving $e^x$, isolate $e^{\ldots}$ and then take $\ln$ of both sides.

When a question involves sums of logarithms, use the rules from Formulae and Tables: $\ln a + \ln b = \ln(ab)$, and $\ln a - \ln b = \ln\frac{a}{b}$. Then, if $\ln A = \ln B$, it follows that $A = B$.

Back to the temperatures: $F = 1.8C + 32$ is bijective, and its inverse is $C = \dfrac{F - 32}{1.8}$. Check: a forecast of $68^\circ$F gives $\dfrac{68-32}{1.8} = \dfrac{36}{1.8} = 20^\circ$C. ✓

Summary