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