If f(x) = log(1 + x) - frac{x}{1 + x}, then t | vedclass

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(D) Step $1$: Find the derivative $f'(x)$.$f(x) = \log(1 + x) - \frac{x}{1 + x}$$f'(x) = \frac{1}{1 + x} - \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac

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$(0, \infty), (-1, 0)$

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(D) Step $1$: Find the derivative $f'(x)$.$f(x) = \log(1 + x) - \frac{x}{1 + x}$$f'(x) = \frac{1}{1 + x} - \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2}$$f'(x) = \frac{(1 + x) - 1}{(1 + x)^2} = \frac{x}{(1 + x)^2}$Step $2$: Determine the intervals for monotonicity.For $f(x)$ to be increasing, $f'(x) > 0 \implies \frac{x}{(1 + x)^2} > 0$. Since $(1 + x)^2 > 0$ for all $x 
eq -1$, $f'(x) > 0$ when $x > 0$.For $f(x)$ to be decreasing, $f'(x)  -1$ due to the domain of $\log(1+x)$).Thus, $f(x)$ is increasing on $(0, \infty)$ and decreasing on $(-1, 0)$.

If $f(x) = \log(1 + x) - \frac{x}{1 + x}$, then the values of $x$ for which $f(x)$ is monotonically increasing and monotonically decreasing are respectively.....

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