Let `f` be a real function defined by `f(x)=sqrt(x-1)` . Find `(fofof)(x)dot` Also, show that `fof!=f^2` .

Explanation

`f(x)=sqrt{x-1}`
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(f o f o f)(x)=f o f(f(x))=f o f(sqrt{x-1})=f(sqrt{sqrt{x-1}-1})`
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=(sqrt{sqrt{sqrt{x-1}-1}-1})`
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Now, f^{2}(x)=(sqrt{x-1})^{2}=x-1`
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f o f(x)=f(sqrt{x-1})=sqrt{sqrt{x-1}-1}`
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therefore f^{2} ne fof`
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