Question

Let f: R → R and g: R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

Answer 1

Given as f: R → R and g: R → R.

Therefore, the domains of f and g are the same.

Consider that (fog) (x) = f(g(x)) 

= f(x + 1) = (x + 1)² 

= x² + 1 + 2

Again consider that (gof)(x) = g(f(x)) 

= g(x²) = x² + 1

Therefore, fog ≠ gof