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

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(C) 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}}{[

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decreasing on $(0, \infty)$.

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

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

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