f: R rightarrow R and g:[0, infty) rightar | vedclass

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(A) Given $ f(x) = x^2 $ and $ g(x) = \sqrt{x} $.Check option $ (A) $: $ f \circ g(-4) = f(g(-4)) $. Since the domain of $ g $ is $ [0, \infty) $,$ g(

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

Vedclass · Answer

(A) Given $ f(x) = x^2 $ and $ g(x) = \sqrt{x} $.Check option $ (A) $: $ f \circ g(-4) = f(g(-4)) $. Since the domain of $ g $ is $ [0, \infty) $,$ g(-4) $ is not defined. Thus,$ f \circ g(-4) $ is not defined. This statement is not true.Check option $ (B) $: $ g \circ f(-2) = g(f(-2)) = g((-2)^2) = g(4) = \sqrt{4} = 2 $. This is true.Check option $ (C) $: $ g \circ f(4) = g(f(4)) = g(4^2) = g(16) = \sqrt{16} = 4 $. This is true.Check option $ (D) $: $ f \circ g(2) = f(g(2)) = f(\sqrt{2}) = (\sqrt{2})^2 = 2 $. This is true.Therefore,the statement that is not true is $ (A) $.

$ 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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