Composition of Functions

Composition of functions. A method of combining a function $f:A\to B$ with a function $g:B\to C$ is composition, defined as $(f\circ g)(x)=f[g(x)]$, read '$f$ composed with $g$' (Fig. 6.35).

Note: (1) The domain of $g\circ f$ is the set of all $x$ in $A$ such that $f(x)$ is in $B$. The range of $g\circ f$ is the set of all $g[f(x)]$ in $C$ such that $f(x)$ is in $B$. (2) Domain of $g\circ f\subseteq$ domain of $f$, and range of $g\circ f\subseteq$ range of $g$.

Illustration: A cow produces 4 litres of milk in a day. Then $x$ cows produce $4x$ litres of milk in a day, given by the function $f(x)=4x='y'$. The price of one litre of milk is Rs. 50, so the price of $y$ litres is Rs. $50y$, given by another function $g(y)=50y$. Now a function $h(x)$ gives the money earned from $x$ cows in a day as a composite function of $f$ and $g$: $h(x)=(g\circ f)(x)=g[f(x)]=g(4x)=50(4x)=200x$.

Ex. 3: If $f(x)=\dfrac{2}{x+5}$ and $g(x)=x^2-1$, find (i) $(f\circ g)(x)$, (ii) $(g\circ f)(3)$.

Solution: (i) Since $(f\circ g)(x)=f[g(x)]$ and $f(x)=\dfrac{2}{x+5}$, replace $x$ in $f(x)$ by $g(x)$: $(f\circ g)(x)=\dfrac{2}{g(x)+5}=\dfrac{2}{x^2-1+5}=\dfrac{2}{x^2+4}$.

(ii) Since $(g\circ f)(x)=g[f(x)]$ and $g(x)=x^2-1$, replace $x$ by $f(x)$: $(g\circ f)(x)=[f(x)]^2-1=\left(\dfrac{2}{x+5}\right)^2-1$. Now let $x=3$: $(g\circ f)(3)=\left(\dfrac{2}{8}\right)^2-1=\left(\dfrac14\right)^2-1=\dfrac{1}{16}-1=\dfrac{1-16}{16}=-\dfrac{15}{16}$.

Ex. 4: If $f(x)=x^2$, $g(x)=x+5$, and $h(x)=\dfrac1x,\ x\ne0$, find $(g\circ f\circ h)(x)$.

Solution: $(g\circ f\circ h)(x)=g\{f[h(x)]\}=g\left[f\left(\dfrac1x\right)\right]=g\left[\left(\dfrac1x\right)^2\right]=\left(\dfrac1x\right)^2+5=\dfrac{1}{x^2}+5$.

Ex. 5: If $h(x)=(x-5)^2$, find the functions $f$ and $g$ such that $h=f\circ g$. …