Question

Find the interval in which the following function are increasing or decreasing \[f\left( x \right) = \log\left( 2 + x \right) - \frac{2x}{2 + x}, x \in R\] ?

Answer 1

\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]

\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]

\[f\left( x \right) = \log\left( 2 + x \right) - \frac{2x}{2 + x}, x \in R\]

\[f'\left( x \right) = \frac{1}{\left( 2 + x \right)} - \frac{\left[ \left( 2 + x \right)2 - 2x \right]}{\left( 2 + x \right)^2}\]

\[ = \frac{\left( 2 + x \right) - \left[ 4 + 2x - 2x \right]}{\left( 2 + x \right)^2}\]

\[ = \frac{2 + x - 4}{\left( 2 + x \right)^2}\]

\[ = \frac{\left( x - 2 \right)}{\left( 2 + x \right)^2}, x \neq - 2\]

\[\text{ Here, x = 2 is the critical point}.\]

\[\text { The possible intervals are }\left( - \infty , 2 \right)\text { and }\left( 2, \infty \right). .....(1)\]

\[\text { For f(x) to be increasing, we must have }\]

\[f'\left( x \right) > 0\]

\[ \Rightarrow \frac{\left( x - 2 \right)}{\left( 2 + x \right)^2} > 0\]

\[ \Rightarrow x - 2 > 0, x \neq - 2\]

\[ \Rightarrow x > 2\]

\[ \Rightarrow x \in \left( 2, \infty \right) \left[ \text { From eq. } (1) \right]\]

\[\text{ So,f(x)is increasing on x }\in \left( 2, \infty \right) .\]

\[\text { For f(x) to be decreasing, we must have }\]

\[f'\left( x \right) < 0\]

\[ \Rightarrow \frac{\left( x - 2 \right)}{\left( 2 + x \right)^2} < 0\]

\[ \Rightarrow x - 2 < 0, x \neq - 2\]

\[ \Rightarrow x < 2\]

\[ \Rightarrow x \in \left( - \infty , 2 \right) \left[ \text { From eq.} (1) \right]\]

\[\text { So,f(x)is decreasing on x }\in \left( - \infty , 2 \right) .\]