Let f(x) = x^2, x in R. For any A subseteq R, | vedclass

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(D) Given $f(x) = x^2$ and $S = [0, 4]$.First,find $g(S) = \{x \in R : x^2 \in [0, 4]\} = \{x \in R : x^2 \leq 4\} = [-2, 2]$.Now,evaluate $f(g(S)) =

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$g(f(S)) = g(S)$

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(D) Given $f(x) = x^2$ and $S = [0, 4]$.First,find $g(S) = \{x \in R : x^2 \in [0, 4]\} = \{x \in R : x^2 \leq 4\} = [-2, 2]$.Now,evaluate $f(g(S)) = f([-2, 2]) = [0, 4] = S$.Evaluate $f(S) = f([0, 4]) = [0, 16]$.Check option $A$: $f(g(S)) = [0, 4]$ and $f(S) = [0, 16]$,so $f(g(S)) 
eq f(S)$. This is true.Check option $B$: $f(g(S)) = [0, 4] = S$. This is true.Check option $C$: $g(f(S)) = g([0, 16]) = \{x \in R : x^2 \in [0, 16]\} = [-4, 4]$. Since $[-4, 4] 
eq [0, 4]$,$g(f(S)) 
eq S$. This is true.Check option $D$: $g(f(S)) = [-4, 4]$ and $g(S) = [-2, 2]$. Thus,$g(f(S)) 
eq g(S)$. Therefore,the statement $g(f(S)) = g(S)$ is not true.

Let $f(x) = x^2, x \in R$. For any $A \subseteq R$,define $g(A) = \{x \in R : f(x) \in A\}$. If $S = [0, 4]$,then which one of the following statements is not true?

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