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(A) Given $f(x) = x^2$ and $g(x) = \sqrt{x}$.For $(f \circ g)(x) = f(g(x))$,the domain is restricted by the domain of $g(x)$,which is $[0, \infty)$.Th

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$(f \circ g)(-4)=4$

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(A) Given $f(x) = x^2$ and $g(x) = \sqrt{x}$.For $(f \circ g)(x) = f(g(x))$,the domain is restricted by the domain of $g(x)$,which is $[0, \infty)$.Thus,$(f \circ g)(x) = (\sqrt{x})^2 = x$ for $x \ge 0$.For $(g \circ f)(x) = g(f(x)) = \sqrt{x^2} = |x|$.Evaluating the options:$A$: $(f \circ g)(-4)$ is undefined because $-4$ is not in the domain of $g(x) = [0, \infty)$.$B$: $(f \circ g)(2) = 2$.$C$: $(g \circ f)(-2) = |-2| = 2$.$D$: $(g \circ f)(4) = |4| = 4$.Since $(f \circ g)(-4)$ is undefined,option $A$ is not true.

$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

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