Inverse functions and their graphsEdexcel International A Level Maths: Revision notes

The inverse function $f^{-1}$ reverses $f$: if $f(a)=b$ then $f^{-1}(b)=a$. It exists only if $f$ is one-one. A many-one function, such as $g(x)=x^2-6x+5$ on $\mathbb{R}$ where $g(2)=g(4)$, has no inverse, but restricting the domain (for example to $x\geqslant3$) can fix that. The two functions undo each other: $f^{-1}f(x)=ff^{-1}(x)=x.$

1. Write $y=f(x)$.
2. Make $x$ the subject (for a fraction, multiply out and collect the $x$ terms; for a square, complete the square).
3. Swap the letters, so $f^{-1}(x)$ is a function of $x$. Example: $y=\frac{2x+1}{x-3}\Rightarrow yx-3y=2x+1\Rightarrow x(y-2)=3y+1$, so $f^{-1}(x)=\frac{3x+1}{x-2}$. For $g(x)=x^2-6x+5=(x-3)^2-4$ on $x\geqslant3$: $x=3+\sqrt{y+4}$, so $g^{-1}(x)=3+\sqrt{x+4}$. Choose the root that fits the domain.

Domain and range swap: the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. For $f(x)=\frac{2x+1}{x-3}=2+\frac{7}{x-3}$, the range is $f(x)\neq2$, so $f^{-1}$ has domain $x\neq2$. For $f(x)=2\mathrm{e}^x-5$, the range is $f(x)>-5$, so $f^{-1}(x)=\ln\frac{x+5}{2}$ has domain $x>-5$.

The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$: a point $(a,b)$ on one graph becomes $(b,a)$ on the other. For $f(x)=2\mathrm{e}^x-5$, $(0,-3)$ lies on $f$, so $(-3,0)$ lies on $f^{-1}$. If $f$ is increasing, the graphs of $f$ and $f^{-1}$ meet on $y=x$, so $f(x)=f^{-1}(x)$ can be solved as $f(x)=x$. For $g(x)=x^2-4x$ ($x\geqslant2$) this gives $x=5$.

If $f^{-1}=f$, the function is self-inverse, so $\mathrm{ff}(x)=x$. The graph is symmetrical about $y=x$. For example $f(x)=\frac{x+3}{x-1}$ gives $x=\frac{y+3}{y-1}$, which is the same rule. The simplest example is $f(x)=\frac1x$.