Let f : R → R : f(x) = x2 and g : R → R : g(x) = (x + 1).
Show that (g o f) ≠ (f o g).
To prove: (g o f) ≠ (f o g)
Formula used: (i) g o f = g(f(x))
(ii) f o g = f(g(x))
Given: (i) f : R → R : f(x) = x2
(ii) g : R → R : g(x) = (x + 1)
Proof: We have,
g o f = g(f(x)) = g(x2) = ( x2 + 1 )
f o g = f(g(x)) = g(x+1) = [ (x+1)2 + 1 ] = x2 + 2x + 2
From the above two equation we can say that (g o f) ≠ (f o g)
Hence Proved