2.7 Composite and inverse functionsIB Maths: Applications and Interpretation HL: Revision notes

A composite function applies one function after another. $(f\circ g)(x)=f(g(x))$ means: apply $g$ first, then $f$ to the result. Example: $f(x)=2x+3$ and $g(x)=x^2$. Then $(f\circ g)(x)=f(x^2)=2x^2+3$, while $(g\circ f)(x)=(2x+3)^2$. In general $f\circ g\neq g\circ f$, so the order matters. The output of $g$ must lie in the domain of $f$. To solve $(f\circ g)(x)=k$, write out the composite and solve: $2x^2+3=11$ gives $x=\pm2$.

In a model, each function is one step of a process, and the composite links the start to the end. If a taxi fare in EUR is $f(d)=3+1.2d$ and conversion to AED is $g(x)=4x$, then the fare in AED is $(g\circ f)(d)=4(3+1.2d)=12+4.8d$. Check the order by asking what goes into the first function. Here the distance $d$ goes into $f$, so $f$ is the inner function. Use your GDC to evaluate a composite at a value, for example $(R\circ V)(6)$, and interpret the result with units.

The inverse function $f^{-1}$ reverses $f$: if $f(a)=b$ then $f^{-1}(b)=a$. It exists only if $f$ is one-to-one (each output comes from exactly one input). Then $(f\circ f^{-1})(x)=(f^{-1}\circ f)(x)=x.$ To find $f^{-1}$: write $y=f(x)$, rearrange to make $x$ the subject, then replace $y$ by $x$. Example: $f(x)=2x+3$ gives $x=\frac{y-3}{2}$, so $f^{-1}(x)=\frac{x-3}{2}$. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. Notation: $f^{-1}(x)$ is not $\frac{1}{f(x)}$.

A many-to-one function has no inverse, but restricting its domain can fix this. Example: $f(x)=(x-3)^2-2$ has $f(2)=f(4)=-1$, so it has no inverse on $\mathbb{R}$. Restrict to $x\geq3$ (or to $x\leq3$) and it is one-to-one. For $x\geq3$: $(x-3)^2=y+2$ and $x-3\geq0$, so $f^{-1}(x)=3+\sqrt{x+2}$, with domain $x\geq-2$. For $x\leq3$ you take the negative root, $f^{-1}(x)=3-\sqrt{x+2}$. The chosen domain of $f$ decides which root to use.

An inverse answers the reverse question. If $f(d)=3+1.2d$ gives a fare from a distance, $f^{-1}(x)=\frac{x-3}{1.2}$ gives the distance for a fare. If $V(d)=\frac{4000}{1+19e^{-0.5d}}$ gives visitors after $d$ days, then solving for $d$ gives $V^{-1}(v)=2\ln\frac{19v}{4000-v}$, the time needed to reach $v$ visitors; $V^{-1}(3000)=8.09$ days. Rearrange step by step, and use logarithms to undo an exponential. For a single value you may instead solve $V(d)=3000$ on your GDC. State the units in the answer.