2.2 Functions, domain, range and inverseIB Maths: Analysis and Approaches HL: Revision notes

A function is a rule that assigns to each input exactly one output. The set of allowed inputs is the domain; the set of outputs actually produced is the range. The graph of $f$ is the set of points $(x, f(x))$.

• A relation is a function if every vertical line meets its graph at most once (the vertical line test).
• Unless stated otherwise, the domain is the largest set of real numbers for which the rule makes sense. For $\sqrt{x-3}$ you need $x \ge 3$; for $\frac{1}{x+2}$ you need $x \ne -2$.
• To find a range, think about the smallest and largest outputs: for $f(x) = 2\sqrt{x-3} + 1$, $\sqrt{x-3} \ge 0$, so $f(x) \ge 1$.

Function notation names the rule and the input: $f(x)$, $v(t)$ for velocity at time $t$, $C(n)$ for the cost of $n$ items. $C(12)$ means substitute $n = 12$; it does not mean $C \times 12$.

A function can be a mathematical model of a real situation. Then the domain must make sense in context: a taxi fare $C(d) = 3.5 + 1.8d$ only applies for distances the taxi drives, e.g. $0 < d \le 40$. The range follows from the domain: here $3.5 < C(d) \le 75.5$.

For an increasing or decreasing linear model, the range on an interval comes from the values at the two ends of the domain. Take care with strict ($<$) and non-strict ($\le$) inequalities at each end.

An inverse function $f^{-1}$ reverses the effect of $f$: if $f(a) = b$ then $f^{-1}(b) = a$.

So solving $f(x) = 10$ is the same as finding $f^{-1}(10)$. For the taxi, a fare of 48.50 dollars means $C(d) = 48.5$, so $d = C^{-1}(48.5) = \frac{48.5 - 3.5}{1.8} = 25$ km.

To find $f^{-1}(x)$ for a simple function, write $y = f(x)$, swap $x$ and $y$, and rearrange for $y$. For $F(c) = 1.8c + 32$: $x = 1.8y + 32$ gives $F^{-1}(x) = \frac{x - 32}{1.8}$.

Swapping inputs and outputs swaps the coordinates of every point: $(a, b)$ on $y = f(x)$ becomes $(b, a)$ on $y = f^{-1}(x)$. So the graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$.

Because of this swap:

• the domain of $f^{-1}$ is the range of $f$, and
• the range of $f^{-1}$ is the domain of $f$.

Points where $y = f(x)$ meets $y = x$ are also on $y = f^{-1}(x)$. For $F(x) = 1.8x + 32$, $F(x) = x$ gives $x = -40$: $-40\,^{\circ}\mathrm{C} = -40\,^{\circ}\mathrm{F}$.

An inverse exists only if $f$ is one-to-one: every output comes from exactly one input. Otherwise the 'inverse' would send one input to two outputs and would not be a function.

Test: every horizontal line meets the graph at most once (the horizontal line test). To show a function is not one-to-one, give two different inputs with the same output: for $g(x) = (x-2)^2 + 1$, $g(0) = g(4) = 5$.

A function that is not one-to-one can be given an inverse by restricting its domain. For $g$, restricting to $x \ge 2$ (from the vertex) keeps only the right-hand half, which is one-to-one. Then when solving $g(x) = 10$ you must reject the solution $x = -1$, which is outside the restricted domain.