Composite and inverse functionsEdexcel A-Level Maths: Revision notes

A function maps each input in its domain to exactly one output; the set of outputs is the range. We write $f:x\mapsto3x-2$ or $f(x)=3x-2$. A one-one function gives each output from exactly one input (e.g. $f(x)=x^3$). A many-one function gives some outputs from more than one input (e.g. $f(x)=x^2$, $x\in\mathbb{R}$, where $f(2)=f(-2)$). A one-many relation such as $y^2=x$ is not a function. To find a range, complete the square or sketch. For $f(x)=x^2-4x+7$, $x\geq2$: $f(x)=(x-2)^2+3$, with minimum $3$ at $x=2$, so the range is $f(x)\geq3$.

$fg(x)$ means $f(g(x))$: do $g$ first, then $f$. With $f(x)=3x-2$ and $g(x)=x^2+1$: $fg(x)=3(x^2+1)-2=3x^2+1,\qquad gf(x)=(3x-2)^2+1=9x^2-12x+5.$ So $fg\neq gf$ in general. $fg$ is only defined where the range of $g$ lies inside the domain of $f$. For $f(x)=\sqrt{x}$ ($x\geq0$) and $g(x)=x-4$, $fg(x)=\sqrt{x-4}$ needs $x\geq4$. Repeated application is written $f^2(x)=f(f(x))$.

The inverse function $f^{-1}$ reverses $f$, and exists only if $f$ is one-one (restrict the domain if needed). Then $f^{-1}f(x)=ff^{-1}(x)=x.$ The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. Method: write $y=f(x)$, make $x$ the subject, then replace $y$ by $x$. Example: $f(x)=x^2-4x+7$, $x\geq2$. $(x-2)^2=y-3$, and as $x\geq2$ we take the positive root: $f^{-1}(x)=2+\sqrt{x-3}$, domain $x\geq3$. Example: $h(x)=\frac{2x+1}{x-3}$. $y(x-3)=2x+1\Rightarrow x(y-2)=3y+1$, so $h^{-1}(x)=\frac{3x+1}{x-2}$, $x\neq2$.

The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$, because inputs and outputs swap: if $(a,b)$ is on $y=f(x)$ then $(b,a)$ is on $y=f^{-1}(x)$. Where an increasing function meets its inverse, the meeting point lies on $y=x$, so $f(x)=f^{-1}(x)$ can then be solved as $f(x)=x$. For $f(x)=x^2-4x+7$, $x\geq2$, the point $(2,3)$ gives $(3,2)$ on $f^{-1}$. A vertical asymptote of $f$ becomes a horizontal asymptote of $f^{-1}$: for $h$, $x=3$ becomes $y=3$ for $h^{-1}$, and the horizontal asymptote $y=2$ becomes the vertical asymptote $x=2$.

• State the domain/range of a composite: check the inner function's range fits the outer function's domain. For $f(x)=e^{2x}-3$ and $g(x)=\ln(x+4)$, the range of $f$ is $f(x)>-3$ and the domain of $g$ is $x>-4$, so $gf$ is defined for all real $x$.
• Simplify with inverse pairs: $e^{\ln u}=u$, so $fg(x)=e^{2\ln(x+4)}-3=(x+4)^2-3=x^2+8x+13$.
• Solve equations: $fg(x)=49$ with $fg(x)=3x^2+1$ gives $x=\pm4$.
• Always give the domain of an inverse. It is the range of the original function.