The function f(x) = frac{log(pi + x)}{log(e + | vedclass

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Question

(B) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.Taking the derivative using the quotient rule:$f'(x) = \frac{\frac{1}{\pi + x} \cdot \log(e + x)

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$A$ decreasing function on $(0, \infty)$.

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(B) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.Taking the derivative using the quotient rule:$f'(x) = \frac{\frac{1}{\pi + x} \cdot \log(e + x) - \frac{1}{e + x} \cdot \log(\pi + x)}{(\log(e + x))^2}$.$f'(x) = \frac{(e + x)\log(e + x) - (\pi + x)\log(\pi + x)}{(\pi + x)(e + x)(\log(e + x))^2}$.Let $g(t) = t \log t$. Then $g'(t) = 1 + \log t$. For $t > 0$,$g(t)$ is increasing for $t > 1/e$.Since $x > 0$,$e + x > e > 1/e$ and $\pi + x > \pi > 1/e$.Since $\pi > e$,we have $\pi + x > e + x$.Because $g(t)$ is an increasing function for $t > 1/e$,$g(\pi + x) > g(e + x)$.Therefore,$(\pi + x)\log(\pi + x) > (e + x)\log(e + x)$.This implies $(e + x)\log(e + x) - (\pi + x)\log(\pi + x) Thus,$f'(x) Hence,$f(x)$ is a decreasing function on $(0, \infty)$.

The function $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$ is:

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