Let f(x) = x 2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.

Question

Philoid · Accepted Answer

We have, f (x) = x² + x + 1 and g(x) = sin x

Now,

fog(x) = f(g(x)) = f(sin x)

⇒ fog(x) = sin² x + sin x + 1

Again, gof(x) = g(f(x)) = g (x² + x + 1)

⇒ gof(x) = sin(x² + x + 1)

Clearly,

fog ≠ gof