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If f:R -> R, g:R -> R defined as f(x) = ...

If `f:R -> R, g:R -> R` defined as `f(x) = sin x and g(x) = x^2`, then find the value of `(gof)(x) and (fog)(x) `and also prove that `gof != fog`.

Text Solution

Verified by Experts

Let `x in R`
`therefore (fog)x=f{g(x)} " "[because g(x)=x^(2)]`
`=f{x^(2)}=sinx^(2)" ... (i)"`
`[because f(x)=sinx]`
and (gof) `x=g{f(x)}`
`=g(sinx)" "[becausef(x)=sinx]`
`=sin^(2)x" ... (ii)"`
`[because g(x)=x^(2)]`
From Eqs. (i) and (ii), we get (fog) x `ne`(gof)x, `AAx inR` Hence, fog `ne` gof
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