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

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(B) Given function: $f(x) = \frac{1}{x + 1} - \log(1 + x)$.To determine if the function is increasing or decreasing,we find its derivative $f'(x)$ wit

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$A$ decreasing function

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(B) Given function: $f(x) = \frac{1}{x + 1} - \log(1 + x)$.To determine if the function is increasing or decreasing,we find its derivative $f'(x)$ with respect to $x$.$f'(x) = \frac{d}{dx} \left( \frac{1}{x + 1} \right) - \frac{d}{dx} (\log(1 + x))$$f'(x) = -\frac{1}{(x + 1)^2} - \frac{1}{1 + x}$.We can factor out the negative sign:$f'(x) = - \left[ \frac{1}{(x + 1)^2} + \frac{1}{x + 1} \right]$.Since $x > 0$,both $(x + 1)$ and $(x + 1)^2$ are positive.Therefore,$\frac{1}{(x + 1)^2} + \frac{1}{x + 1} > 0$ for all $x > 0$.This implies that $f'(x)  0$.Since the derivative $f'(x)$ is negative for all $x$ in the given domain,the function $f(x)$ is a decreasing function.

If $f(x) = \frac{1}{x + 1} - \log(1 + x)$,where $x > 0$,then $f$ is:

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