Question

Find the intervals in which f(x) = log (1 + x) − $\frac{x}{1+x}$ is increasing or decreasing.

Answer 1

$$\begin{gathered} f\left(x\right)=\log \left(1+x\right)-\frac{x}{1+x} \\ \text{Domain of}f\left(x\right)\text{is}\left(-1,\infty \right)\text{.} \\ f\text{'}\left(x\right)=\frac{1}{1+x}-\left\{\frac{1+x-x}{{\left(1+x\right)}^2}\right\} \\ =\frac{1}{1+x}-\frac{1}{{\left(1+x\right)}^2} \\ =\frac{x}{{\left(1+x\right)}^2} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{x}{{\left(1+x\right)}^2}>0 \\ \Rightarrow x>0\left[\text{∵}{\left(1+x\right)}^2\text{>0, Domain:}\left(-1,\infty \right)\right] \\ \Rightarrow x\in \left(0,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(0,\infty \right). \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \frac{x}{{\left(1+x\right)}^2}<0 \\ \Rightarrow x<0\left[\because {\left(1+x\right)}^2\text{>0, Domain:}\left(-1,\infty \right)\right] \\ \Rightarrow x\in \left(-1,0\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-1,0\right)\text{.}\mathit{} \end{gathered}$$