Question
Find the intervals in which $\text{f}(\text{x})=\log(1+\text{x})-\frac{\text{x}}{1+\text{x}}$ is increasing or decreasing.
✓
Answer
$\text{f}(\text{x})=\log(1+\text{x})-\frac{\text{x}}{1+\text{x}}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}}-\Big(\frac{(1+\text{x})-\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{1}{1+\text{x}}-\frac{1}{(1+\text{x})^2}$
$=\frac{\text{x}}{(1+\text{x})^2}$
Critical points,
$\text{f}'(\text{x})=0$
$\Rightarrow\frac{\text{x}}{(1+\text{x})^2}=0$
$\Rightarrow\text{x}=0,-1 $
Clearly, f'(x) > 0 if x > 0 and f'(x) < 0 if -1 < x < 0 or x < -1
Hence, f(x) increases in $(0,\infty),$ decreases in $(-\infty,-1)\cup(-1,0).$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}}-\Big(\frac{(1+\text{x})-\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{1}{1+\text{x}}-\frac{1}{(1+\text{x})^2}$
$=\frac{\text{x}}{(1+\text{x})^2}$
Critical points,
$\text{f}'(\text{x})=0$
$\Rightarrow\frac{\text{x}}{(1+\text{x})^2}=0$
$\Rightarrow\text{x}=0,-1 $
Clearly, f'(x) > 0 if x > 0 and f'(x) < 0 if -1 < x < 0 or x < -1
Hence, f(x) increases in $(0,\infty),$ decreases in $(-\infty,-1)\cup(-1,0).$
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