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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{2}\text{x}^4-4\text{x}^3-45\text{x}^2+51$
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Answer

$\text{f}(\text{x})=\frac{3}{2}\text{x}^4-4\text{x}^3-45\text{x}^2+51$
$f'(x) = 6x^3 - 12x^2 - 90x = 6x(x^2 - 2x - 15) = 6x(x - 5)(x + 3)$
Here, $x = -3, x = 0$ and $x = 5$ are the critical points.
The possible intervals are $(-\infty,-3),(-3,0),(0,5)$ and $(5,\infty)\ ....(1)$
For $f(x)$ to be increasing,
we must have$ f'(x) > 0 \Rightarrow 6x(x - 5)(x + 3) > 0$
$[$Since, $6 > 0, 6x(x - 5)(x + 3) > 0 \Rightarrow x(x - 5)(x + 3) > 0] $
$\Rightarrow x(x - 5)(x + 3) > 0$
$\Rightarrow\text{x}\in(-3,0)\cup(5,\infty)$
$[$From eq. $1]$ So, $f(x)$ is increasing on $\text{x}\in(-3,0)\cup(5,\infty).$
For f(x) to be decreasing,
we must have, $f'(x) < 0 \Rightarrow 6x(x - 5)(x + 3) < 0$
$[$Since, $6 > 0, 6x(x - 5)(x + 3) < 0 \Rightarrow x(x - 5)(x + 3) < 0]$
$\Rightarrow x(x - 5)(x + 3) < 0$
$\Rightarrow\text{x}\in(-\infty,-3)\cup(0,5)$
$[$From eq. $1]$So, $f(x)$ is decreasing on
$\text{x}\in(-\infty,-3)\cup(0,5).$

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