Which one of the following functions is monot | vedclass

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(A) To determine if a function is monotonically increasing,we check if its derivative $f'(x) \ge 0$ for all $x$ in its domain.For option $A$: $f'(x) =

Vedclass · Accepted Answer

$f(x) = \log(1+x) - x + \frac{x^2}{2}$

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(A) To determine if a function is monotonically increasing,we check if its derivative $f'(x) \ge 0$ for all $x$ in its domain.For option $A$: $f'(x) = \frac{1}{1+x} - 1 + x = \frac{1 - (1+x) + x(1+x)}{1+x} = \frac{1 - 1 - x + x + x^2}{1+x} = \frac{x^2}{1+x}$. For $x > -1$,$f'(x) \ge 0$,so it is monotonically increasing.For option $B$: $g'(x) = \frac{2}{1+x^2} - 1 = \frac{2 - 1 - x^2}{1+x^2} = \frac{1-x^2}{1+x^2}$. This changes sign at $x = \pm 1$,so it is not monotonically increasing.For option $C$: $h'(x) = -4 \sin x + 1$. This changes sign depending on $\sin x$,so it is not monotonically increasing.For option $D$: $u'(x) = \frac{1}{1+x} - \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{1+x} - \frac{1}{(x+1)^2} = \frac{x+1-1}{(x+1)^2} = \frac{x}{(x+1)^2}$. This changes sign at $x = 0$,so it is not monotonically increasing.Thus,the correct option is $A$.

Which one of the following functions is monotonically increasing in its domain?

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