Inverse Functions

Inverse functions. Let $f:A\to B$ be a one-one and onto function, with $f(x)=y$ for $x\in A$. The inverse function $f^{-1}:B\to A$ is defined by $f^{-1}(y)=x$ if $f(x)=y$ (Fig. 6.36).

Note: (1) Since $f$ is one-one and onto, every $y\in B$ has a unique $x\in A$ with $y=f(x)$, so $f^{-1}$ is itself a well-defined function. (2) If $f$ and $g$ are one-one and onto functions such that $f[g(x)]=x$ for every $x$ in the domain of $g$, and $g[f(x)]=x$ for every $x$ in the domain of $f$, then $g$ is called the inverse of $f$, denoted $f^{-1}$ (read 'f inverse'). That is, $f[g(x)]=g[f(x)]=x$ means $g=f^{-1}$, equivalently $f[f^{-1}(x)]=f^{-1}[f(x)]=x$. (3) $f^{-1}(x)\ne[f(x)]^{-1}$ — these look similar but mean completely different things: $[f(x)]^{-1}=\dfrac{1}{f(x)}$ is the reciprocal of $f(x)$, whereas $f^{-1}(x)$ is the inverse function of $f$. For example, if $f$ is one-one and onto with $f(3)=7$, then $f^{-1}(7)=3$.

Ex. 7: If $f$ is a one-one onto function with $f(x)=9-5x$, find $f^{-1}(-1)$.

Solution: Let $f^{-1}(-1)=m$, so $-1=f(m)$. Therefore $-1=9-5m \implies 5m=9+1=10 \implies m=2$. That is, $f(2)=-1$, so $f^{-1}(-1)=2$.

Ex. 8: Verify that $f(x)=\dfrac{x-5}{8}$ and $g(x)=8x+5$ are inverse functions of each other.

Solution: Since $f(x)=\dfrac{x-5}{8}$, replace $x$ in $f(x)$ with $g(x)$: $f[g(x)]=\dfrac{g(x)-5}{8}=\dfrac{8x+5-5}{8}=\dfrac{8x}{8}=x$. And since $g(x)=8x+5$, replace $x$ in $g(x)$ with $f(x)$: $g[f(x)]=8f(x)+5=8\left(\dfrac{x-5}{8}\right)+5=(x-5)+5=x$. Since $f[g(x)]=x$ and $g[f(x)]=x$, $f$ and $g$ are inverse functions of each other.

Ex. 9: Determine whether $f(x)=\dfrac{2x+1}{x-3}$ has an inverse; if it exists, find it.

Solution: $f^{-1}$ exists only if $f$ is one-one and onto. One-one: consider $f(x_1)=f(x_2)$: $\dfrac{2x_1+1}{x_1-3}=\dfrac{2x_2+1}{x_2-3}$, so $(2x_1+1)(x_2-3)=(2x_2+1)(x_1-3)$, i.e. $2x_1x_2-6x_1+x_2-3=2x_1x_2-6x_2+x_1-3$, i.e. $-6x_1+x_2=-6x_2+x_1$, i.e. $6x_1+x_2=6x_2+x_1$, i.e. $7x_2=7x_1$, i.e. $x_2=x_1$. Hence $f$ is one-one. …