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

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(B) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.Using the quotient rule,$f'(x) = \frac{\ln(e + x) \cdot \frac{1}{\pi + x} - \ln(\pi + x) \cdot \frac{

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Decreasing on $[0, \infty)$

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(B) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.Using the quotient rule,$f'(x) = \frac{\ln(e + x) \cdot \frac{1}{\pi + x} - \ln(\pi + x) \cdot \frac{1}{e + x}}{(\ln(e + x))^2}$.$f'(x) = \frac{(e + x)\ln(e + x) - (\pi + x)\ln(\pi + x)}{(\ln(e + x))^2 (e + x)(\pi + x)}$.Consider the function $g(t) = t \ln(t)$ for $t > 0$. Then $g'(t) = \ln(t) + 1$. For $t > 1/e$,$g'(t) > 0$,so $g(t)$ is an increasing function.Since $\pi > e$,for $x \ge 0$,we have $\pi + x > e + x > e > 1$. Thus,$g(\pi + x) > g(e + x)$,which implies $(\pi + x)\ln(\pi + x) > (e + x)\ln(e + x)$.Therefore,the numerator $(e + x)\ln(e + x) - (\pi + x)\ln(\pi + x) Since the denominator is always positive,$f'(x) Hence,$f(x)$ is decreasing on $[0, \infty)$.

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

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