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

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(B) Given $f(x) = \frac{1}{x + 1} - \log(1 + x)$ for $x > 0$.To determine the nature of the function,we find its derivative $f'(x)$:$f'(x) = \frac{d}{

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Decreasing function

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(B) Given $f(x) = \frac{1}{x + 1} - \log(1 + x)$ for $x > 0$.To determine the nature of the function,we find its derivative $f'(x)$:$f'(x) = \frac{d}{dx} \left( \frac{1}{x + 1} \right) - \frac{d}{dx} (\log(1 + x))$$f'(x) = -\frac{1}{(x + 1)^2} - \frac{1}{1 + x}$$f'(x) = -\left( \frac{1 + (x + 1)}{(x + 1)^2} \right) = -\frac{x + 2}{(x + 1)^2}$Since $x > 0$,we have $x + 2 > 0$ and $(x + 1)^2 > 0$.Therefore,$f'(x) = -\frac{x + 2}{(x + 1)^2}  0$.Since the derivative $f'(x) < 0$ for all $x$ in the domain,the function $f$ is a decreasing function.

If $f(x) = \frac{1}{x + 1} - \log(1 + x)$,where $x > 0$,then what type of function is $f$?

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