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Let f: R → R: f(x) = x2 and g: R → R: g(x) = (x + 1). Show that (g o f) ≠ (f o g).

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It is given that

f(x) = x2 and g(x) = (x + 1)

We know that LHS

gof = gof (x) = g{f(x)} = g{ x2} = x2 + 1

We know that RHS

fog = fog (x) = f{g(x)} = f{x+1} = (x+1)2 = x2 + 2x + 1

Thus, gof ≠ fog

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