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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions5 Marks
Question
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
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Answer

$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
$=\frac{3\text{x}^4-8\text{x}^3-30\text{x}^2+72\text{x}+110}{10}$
$\text{f}'(\text{x})=\frac{12\text{x}^3-24\text{x}^2+60\text{x}+72}{10}$
$=\frac{12}{10}\big(\text{x}^3-2\text{x}^2-5\text{x}+6\big)$
$=\frac{(\text{x}-1)(\text{x}^2-\text{x}-6)}{10}$
$=\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)$
Here, 1, 2, 3 are the Critical points.
The possible intervals are $(-\infty,-2),(-2,-1),(1,3)$ and $(3,\infty).$
For f(x) to be increasing, we must have
$\text{f}'(\text{x})>0$
$\Rightarrow\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)>0$
$\Rightarrow(\text{x}-1)(\text{x}+2)(\text{x}-3)>0$
$\Rightarrow\text{x}\in(-2,1)\cup(3,\infty)$
So, f(x) is increasing on $\text{x}\in(-2,1)\cup(3,\infty).$
For f(x) to be decreasing, we must have
$\text{f}'(\text{x})<0$
$\Rightarrow\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)<0$
$\Rightarrow(\text{x}-1)(\text{x}+2)(\text{x}-3)<0$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(1,3)$
So, f(x) is decreasing on $\text{x}\in(-\infty,-2)\cup(1,3).$

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