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

Answer :

We have, f (x) = x^{2} + x + 1 and g(x) = sin x

Now,

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

⇒ fog(x) = sin^{2} x + sin x + 1

Again, gof(x) = g(f(x)) = g (x^{2} + x + 1)

⇒ gof(x) = sin(x^{2} + x + 1)

Clearly,

fog ≠ gof

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