Question:
Let $f: R \rightarrow R: f(x)=x^{2}$ and $g: R \rightarrow R: g(x)=(x+1)$.
Show that $(g \circ f) \neq(f \circ g) .$
Solution:
To prove: $(g \circ f) \neq(f \circ g)$
Formula used: (i) g o f = g(f(x))
(ii) f o g = f(g(x))
Given: (i) $f: R \rightarrow R: f(x)=x^{2}$
(ii) $g: R \rightarrow R: g(x)=(x+1)$
Proof: We have,
$g \circ f=g(f(x))=g\left(x^{2}\right)=\left(x^{2}+1\right)$
fo $g=f(g(x))=g(x+1)=\left[(x+1)^{2}+1\right]=x^{2}+2 x+2$
From the above two equation we can say that (g o f) ≠ (f o g)
Hence Proved