If f:R to R and g:R to R are given by f(x) = | vedclass

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(D) Given $f(x) = |x|$ and $g(x) = |x|$ for all $x \in R$.We need to find the set of $x$ such that $g(f(x)) \le f(g(x))$.Substitute the functions: $g(

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$R$

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(D) Given $f(x) = |x|$ and $g(x) = |x|$ for all $x \in R$.We need to find the set of $x$ such that $g(f(x)) \le f(g(x))$.Substitute the functions: $g(|x|) \le f(|x|)$.Since $g(x) = |x|$,$g(|x|) = ||x|| = |x|$.Since $f(x) = |x|$,$f(|x|) = ||x|| = |x|$.The inequality becomes $|x| \le |x|$.This inequality holds true for all real numbers $x \in R$ because any real number is less than or equal to itself.Therefore,the set is $R$.

If $f:R \to R$ and $g:R \to R$ are given by $f(x) = |x|$ and $g(x) = |x|$ for each $x \in R$,then $\{ x \in R : g(f(x)) \le f(g(x)) \} = $

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