Question

Find gof and fog if $f:\mathbb{R}\to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$ are given by f(x) = cosx and $g\left( x \right) = 3{{x}^{2}}$. Show that $fog\left( x \right)\ne gof\left(x \right)$.

Answer 1

Hint: In order to prove that the two functions are not equal we need to find a value of x for which the two functions give different values. Hence first construct these functions and hence find a value of x for which fog and gof give different values. Hence prove that the functions are not equal.

Complete step-by-step solution -

Two functions f and g are not equal if Domain of f and Domain of g are different or Range of f and Range of g are different or there exists a value of x for which f(x) is not equal to g(x).
Consider $f\left( x \right)=\cos x$ and $g\left( x \right)=3{{x}^{2}}$
Now we have $fog\left( x \right)=f\left( 3{{x}^{2}} \right)=\cos \left( 3{{x}^{2}} \right)$
We have Domain of fog = R
Range of fog =[-1,1]
$gof\left( x \right)=g\left( \cos x \right)=3{{\cos }^{2}}x$
We have Domain of gof = R
Range of gof = [0,3]
Since Range of gof is not equal to range of fog, the two functions are not equal. Hence we have
$fog\ne gof$.

Note:[1] Finding the range
$\begin{aligned} & -1\le \cos x\le 1 \\\ & \Rightarrow 0\le {{\cos }^{2}}x\le 1 \\\ & \Rightarrow 0\le 3{{\cos }^{2}}x\le 3 \\\ \end{aligned}$
Hence Range of gof(x) is [0,3]
[2] This example serves as a proof that composition of two functions is not commutative, i.e. in general fog(x) is not the same as gof(x).
Graph of fog(x)

Graph of gof(x)

As is evident from the graphs of the two functions, we have the two functions are not equal.