If f(x)=e^x and g(x)=ln(x) for all x in [1, i | vedclass

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(D) Given $f(x)=e^x$ and $g(x)=\ln(x)$ for $x \in [1, \infty)$.We define the composition $(f \circ g)(x) = f(g(x))$.Substituting the functions,we get

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Bijective

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(D) Given $f(x)=e^x$ and $g(x)=\ln(x)$ for $x \in [1, \infty)$.We define the composition $(f \circ g)(x) = f(g(x))$.Substituting the functions,we get $(f \circ g)(x) = e^{\ln(x)}$.Since $e^{\ln(x)} = x$ for all $x > 0$,we have $(f \circ g)(x) = x$ for $x \in [1, \infty)$.The function $h(x) = x$ defined on the domain $[1, \infty)$ is an identity function.An identity function is both one-one (injective) and onto (surjective) on its domain and range.Therefore,the function is bijective.Hence,option $D$ is correct.

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

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