If f: R rightarrow R is defined by f(x)=e^{x} | vedclass

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(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.We calculate the composite function $(g \circ f)(x)

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$g \circ f$ is injective but $g$ is not bijective

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(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.We calculate the composite function $(g \circ f)(x) = g(f(x)) = g(e^{x}) = (e^{x})^{2} = e^{2x}$.For $(g \circ f)(x) = e^{2x}$, if $(g \circ f)(x_{1}) = (g \circ f)(x_{2})$, then $e^{2x_{1}} = e^{2x_{2}}$, which implies $2x_{1} = 2x_{2}$, so $x_{1} = x_{2}$. Thus, $g \circ f$ is injective.However, the range of $g \circ f$ is $(0, \infty)$, which is not equal to the codomain $R$, so $g \circ f$ is not surjective.For $g(x) = x^{2}$, $g(-1) = 1$ and $g(1) = 1$, so $g$ is not injective. Also, the range of $g$ is $[0, \infty)$, so $g$ is not surjective.Therefore, $g \circ f$ is injective but $g$ is not bijective.

If $f: R \rightarrow R$ is defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ is defined by $g(x)=x^{2}$, then the mapping $(g \circ f): R \rightarrow R$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in R$. Which of the following is true?

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