If x 0,then frac{x}{1+x} - log(1+x) | vedclass

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Question

(A) Let $f(x) = \frac{x}{1+x} - \log(1+x)$ for $x > 0$. First,we find the derivative $f'(x)$: $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} - \frac{1}{1+x}

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is less than zero

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(A) Let $f(x) = \frac{x}{1+x} - \log(1+x)$ for $x > 0$. First,we find the derivative $f'(x)$: $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} - \frac{1}{1+x}$ $f'(x) = \frac{1}{(1+x)^2} - \frac{1}{1+x} = \frac{1 - (1+x)}{(1+x)^2} = \frac{-x}{(1+x)^2}$. Since $x > 0$,we have $f'(x)  0$. This implies that $f(x)$ is a strictly decreasing function for $x > 0$. As $x 	o 0^+$,$f(x) 	o \frac{0}{1} - \log(1) = 0$. Since the function starts at $0$ and is strictly decreasing for $x > 0$,it follows that $f(x)  0$.

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

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