If f(x) and g(x) are two real-valued function | vedclass

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(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$ Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functiona

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$\phi$

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(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$ Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functional equation. The solution to this is $h(x) = cx$ for some constant $c$. Given $g(1) = 2$ and $f(2) = 1$,we have $h(1) = f(g(1)) = f(2) = 1$. Since $h(1) = c(1) = 1$,we get $c = 1$. Thus,$h(x) = f(g(x)) = x$. Since $f(g(x)) = x$,$f$ and $g$ are inverse functions of each other. Therefore,$g(f(x)) = x$ for all $x \in R$. The function $g(f(x)) = x$ is a polynomial function,which is continuous everywhere on the set of real numbers $R$. Hence,the set of points where $g(f(x))$ is discontinuous is the empty set,denoted by $\phi$.

If $f(x)$ and $g(x)$ are two real-valued functions such that $f(g(x+y)) = f(g(x)) + f(g(y))$,$g(1) = 2$,and $f(2) = 1$,then the function $g(f(x))$ is discontinuous on the set

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