Invertible And Composite Functions

INVERSE FUNCTION

If f: X$\to$Y be a function defined by y = f(x) such that f is both one – one and onto, then there exists a unique function g: Y$\to$X such that for each y $\in$ Y, g(y) = x if and only if y = f(x). The function g so defined is called the inverse of f and denoted by f ^{– 1}.

f(a) = a₁$\Rightarrow$ f ^{– 1}(a₁) = a

SOME IMPORTANT POINTS:

$\bullet$ The condition for existence of inverse of a function is that the function must be one – one and onto.

$\bullet$ Whenever an inverse function is defined, the range of the original function becomes the domain of the inverse function and domain of the original function becomes the range of the inverse function.

$\bullet$ Note that fof ^–1(x) = f ^-1of(x) = x always and roots of the equation f(x) = f ^–1(x) would always lie on the line y = x.

$\bullet$ f and f ^–1 are symmetric about the line y = x.

Illustration 1. The function ¦: [1, $\infty$) $\to$ [1, $\infty$) is defined by ¦(x)=2^{x(x – 1)}, find $f$ ^{– 1}(x).

Key concept : First check the function for one – one and onto. And if function is one – one and onto then find inverse using the identity $f\left(f^{-1}\left(\text{x}\right)\right)=\text{x}$

Solution: Given, $f$(x) = 2^{x(x – 1)}$\Rightarrow$log $f$(x) = x(x – 1) log_e2

_{$\Rightarrow \frac{1}{f(\text{x})}{f}^{\prime }(\text{x})=(2\text{x}-\text{1})lo{g}_{e}2$.}

$\Rightarrow$ $f$(x) = 2^{x(x – 1)} log_e2 (2x – 1)

Thus $f$(x) is an increasing function in [1, $\infty$), therefore, $f$(x) is a one – on function.

Also range of f(x) is [1,$\infty$) which is equal to co – domain.

Hence the function is also onto.

TO FIND $f$ ^{– 1}(x):

Let f ^{– 1} be the inverse function of f, then by rule of identity $f\left(f^{-1}\left(\text{x}\right)\right)=\text{x}$

$f\left(f^{-1}(\text{x})\right)=2^{f^{-1}(x)\left(f^{-1}(x)-1\right)}$ = x

$\therefore$ $\Rightarrow lo{g}_2\text{x}={\left(f^{-1}(x)\right)}^2-f^{-1}(x)$

\begin{align} {{f}^{-1}}(x)=\dfrac{1\pm \sqrt{1+4lo{{g}_{2}}\text{x}}}{2} \\  but\,{{f}^{-1}}(x)=\dfrac{1+\sqrt{1+4lo{{g}_{2}}\text{x}}}{2} \\ \end{align}

_{\left( \begin{align}  \because 1\le {{f}^{-1}}(\text{x})<\infty \,\therefore 1+4lo{{g}_{2}}\text{x}\ge 1\,\text{ }and\text{ }\,\text{x}\ge 1 \\ Therefore\,\,only\,\,positive\,\,sign\,\,is\,\text{ }acceptable. \\ \end{align} \right)}

_.

COMPOSITE FUNCTION

If $f$: x $\to$ y and g: y $\to$ z then we define the composite function (go$f$): x $\to$ z by (go$f$) (x) $\equiv$ g {f(x)}. To obtain (go$f$)(x), we first take the f-image of an element x$\in$x so that f(x) $\in$y, which is the domain of g(x). Then take g-image of f(x), i.e.g (f(x)) which would be an element of z.

$\bullet$ (g($f$(${\text{x}}_1$))= g(${\text{y}}_2$)=${\text{z}}_3$.

$\bullet$ Range (go$f$) = {${\text{z}}_2,{\text{z}}_3$} But Range (g) = {${\text{z}}_1,{\text{z}}_2,{\text{z}}_3$}

$\bullet$ Clearly domain (gof) = {x : x $\in$ Domain(¦),$f$(x) $\in$ domain(g)}

$\bullet$ Similarly we can define, (fog)x = f(g(x)) and domain (fog) = {x : x $\in$ Domain(g), g(x) $\in$ domain(f)}. In general fog $\ne$ gof.