2.2 Concept of a function, domain, range and inverseIB Maths: Applications and Interpretation SL: Revision notes

A function is a rule that gives exactly one output for each input. We write $f(x)$ for the output when the input is $x$; other letters fit the context, e.g. $v(t)$ for velocity at time $t$, or $C(n)$ for the cost of $n$ items. The set of allowed inputs is the domain, and the set of outputs that result is the range. A function is often a mathematical model of a real situation: $C(n)=12+2.5n$ models a taxi fare of 12 AED plus 2.5 AED per kilometre. $C(18)=57$ means an 18 km journey costs 57 AED. The graph of a function is the set of points $(x, f(x))$, usually drawn as $y=f(x)$.

The domain may be limited by the mathematics or by the context.

• A square root needs a non-negative expression: for $f(x)=\sqrt{2-x}$ we need $2-x\ge0$, so the domain is $x\le2$ and the range is $f(x)\ge0$.
• In context, negative times or distances are not allowed: for $C(n)=12+2.5n$ the domain is $n\ge0$. To find the range, sketch or graph the function with your GDC and read off the lowest and highest output values. Remember to include or exclude end values correctly: use $\ge$ and $\le$ when the value is reached. Example: $g(x)=2x+5$ for $-1\le x\le4$ is increasing, so the range is $g(-1)\le g(x)\le g(4)$, i.e. $3\le g(x)\le13$.

An inverse function reverses or undoes the effect of $f$. We write it $f^{-1}$. If $f$ turns 5 into 11, then $f^{-1}$ turns 11 back into 5. To find $f^{-1}(x)$ algebraically: write $y=f(x)$, swap $x$ and $y$, then rearrange for $y$. Example: $f(x)=2x+5$. Swap: $x=2y+5$, so $y=\frac{x-5}{2}$ and $f^{-1}(x)=\frac{x-5}{2}$. The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y=x$. Solving $f(x)=10$ is the same as finding $f^{-1}(10)$.

An inverse function exists only for a one-to-one function, where each output comes from exactly one input. If two different inputs give the same output, the inverse would not know which one to return. Example: $f(x)=x^2$ is not one-to-one because $f(-2)=f(2)=4$. If the domain is restricted to $x\ge0$, it is one-to-one and has an inverse, $f^{-1}(x)=\sqrt{x}$. The inputs and outputs swap roles, so:

• the domain of $f^{-1}$ is the range of $f$;
• the range of $f^{-1}$ is the domain of $f$. In the ball-throwing model $h(t)=-5t^2+20t+25$, the domain must be restricted to $2\le t\le5$ before $h^{-1}$ can give a unique time for each height.

In a model, $f^{-1}$ answers the reverse question. For the taxi fare $C(n)=12+2.5n$, $C(18)=57$ asks 'what does 18 km cost?', while $C^{-1}(100)$ asks 'how far can I travel for 100 AED?'. Solve $12+2.5n=100$ to get $n=35.2$ km. Always state what the answer means and give the units. When two inputs give the same output (as with a ball at the same height on the way up and down), use the domain to choose the correct one.