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Let R be the set of all real numbers let `f: R to R : f(x) = sin x ` and `g : R to R : g (x) =x^(2) .` Prove that g o f `ne ` f o g

Answer» Let x be an arbitrary real number . Then
(g o g ) (x) = g { f(x) } =g (sin x) `=(sin x)^(2)`
(f o g) (x) = f { g (x) } =` f (x^(2)) =sin x^(2)`
Clearly `(sin x)^(2) ne sin x^(2)`
Hence g o g `ne ` g o f


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