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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions4 Marks
Question
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\log(2+\text{x})-\frac{2\text{x}}{2+\text{x}},\text{x}\in\text{R}$
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Answer

$\text{f}(\text{x})=\log(2+\text{x})-\frac{2\text{x}}{2+\text{x}},\text{x}\in\text{R}$

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

$=\frac{(2+\text{x})-[4+2\text{x}-2\text{x}]}{(2+\text{x})^2}$

$=\frac{2+\text{x}-4}{(2+\text{x})^2}$

$=\frac{(\text{x}-2)}{(2+\text{x})^2},\text{x}\neq-2$

Here, x = 2 is the critical point.

The possible intervals are $(-\infty,2)$ and $(2,\infty)\ ....(1)$

For f(x) to be increasing, we must have,

$\text{f}'(\text{x})>0$

$\Rightarrow\frac{(\text{x}-2)}{(2+\text{x})^2}>0$

$\Rightarrow\text{x}-2>0,\text{x}\neq-2$

$\Rightarrow\text{x}>2$

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

So, f(x) is increasing on $\text{x}\in(2,\infty).$

For f(x) to be decreasing, we must have,

$\text{f}'(\text{x})<0$

$\Rightarrow\frac{(\text{x}-2)}{(2+\text{x})^2}<0$

$\Rightarrow\text{x}-2<0,\text{x}\neq-2$

$\Rightarrow\text{x}<2$

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

So, f(x) is decreasing on $\text{x}\in(-\infty,2).$

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