Composite and inverse functionsAQA A-Level Maths: Revision notes

A composite function applies one function after another. $fg(x)$ means $f(g(x))$: apply $g$ first, then $f$. For $f(x)=3x-2$ and $g(x)=x^2+1$: $fg(x)=3(x^2+1)-2=3x^2+1,\qquad gf(x)=(3x-2)^2+1=9x^2-12x+5.$ In general $fg\ne gf$. The domain of $fg$ is the set of $x$ in the domain of $g$ for which $g(x)$ lies in the domain of $f$.

Take $f(x)=\sqrt{x+5}$ for $x\ge-5$ and $g(x)=x^2-4$. Then $gf(x)=x+1$, but its domain is still $x\ge-5$ because $f$ only accepts those inputs. Its range is $gf(x)\ge-4$, not all real numbers. Always find the composite first, then state domain from the inner function and range from the values the whole expression can take.

The inverse $f^{-1}$ undoes $f$: $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$. It exists only if $f$ is one-to-one (each output comes from exactly one input); a many-to-one function needs its domain restricted first. Method: write $y=f(x)$, rearrange to make $x$ the subject, then replace $y$ by $x$. Example: $h(x)=\frac{2x+1}{x-3}$. Then $y(x-3)=2x+1$, so $x(y-2)=3y+1$ and $h^{-1}(x)=\frac{3x+1}{x-2}$.

The domain and range swap: the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. Example: $f(x)=(x-2)^2+3$ for $x\ge2$ has range $f(x)\ge3$. Then $f^{-1}(x)=2+\sqrt{x-3}$ with domain $x\ge3$ and range $f^{-1}(x)\ge2$. Use the positive root because the original domain was $x\ge2$.

The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$, because swapping $x$ and $y$ swaps the coordinates of every point. If $f$ is increasing, the graphs of $f$ and $f^{-1}$ can only meet on $y=x$, so solve $f(x)=x$. For $f(x)=\sqrt{x+5}$ this gives $x^2-x-5=0$, so $x=\frac{1+\sqrt{21}}{2}$ (reject the negative root).