Prove that

Question:
Let $f: R \rightarrow R$ and $g: R \rightarrow R$ defined by $f(x)=x^{2}$ and $g(x)=(x+1)$. Show that $g \circ f \neq f \circ g$.

Solution:

To prove: $\mathrm{g}$ of $\mathrm{f} \neq \mathrm{fog}$

Formula used: (i) $f \circ g=f(g(x))$

(ii) $g \circ f=g(f(x))$

Given: (i) $f: R \rightarrow R: f(x)=x^{2}$

(ii) $g: R \rightarrow R: g(x)=(x+1)$

We have,

$f \circ g=f(g(x))=f(x+7)$

$f \circ g=(x+7)^{2}=x^{2}+14 x+49$

$g \circ f=g(f(x))=g\left(x^{2}\right)$

$g \circ f=\left(x^{2}+1\right)=x^{2}+1$

Clearly $\mathrm{g}$ o $\mathrm{f} \neq \mathrm{fog}$

Hence Proved