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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Find the intervals in which the function f given by $f(x) = 2x^2- 3x^2 - 36x + 7$ is $(a)$ strictly increasing, $(b)$ strictly decreasing.
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Answer

Given: $\text{f}\text{(x)} = 2\text{x}^3 - 3\text{x}^2 - 36\text{x} + 7 $ $\Rightarrow\ \text{f}'\text{(x)} = 6\text{x}^2 - 6\text{x} - 36 = 6(\text{x}^2 - \text{x} - 6)$$\Rightarrow\ \text{f}'\text{(x)} = 6(\text{x} + 2)(\text{x}-3)\dots\dots\dots\text {(i)}$
$\text{Now }6(\text{x} + 2)(\text{x} - 3) = 0 \ \Rightarrow\ \text{x} + 2 = 0 \text{ or } \text{x} - 3 = 0$
$\Rightarrow \ \text{x} = -2\text{ or } \text{x} = 3$
Therefore, we have sub-intervals are $(-\infty,-2),(-2,3)\text{ and }(3,\infty).$
For interval $(-\infty,\ -2),$
taking$ x = -3 ($say$)$, from eq.$ (i), f'(x) = (+) (-) (-) = (+) > 0$
 
Therefore, $f$ is strictly increasing in $(-\infty,\ -2).$
For interval $(-2,\ 3),$
taking $x = 2($say$),$ from eq. $(i), f'(x) = (+) (+) (-) = (-) < 0$
 
Therefore, $f$ is strictly decreasing in $(-2, 3).$
For interval $(3,\ \infty),$
taking $x = 4 ($say$), $ from eq. $(i), f'(x) = (+) (+) (+) = (+) > 0$
 
Therefore,$f$ is strictly increasing in $(3,\ \infty).$
Hence,
  1. $f$ is is strictly increasing in $(- \infty,\ -2)\text{ and} (,\ \infty).$
  2. $f$ is strictly decreasing in $(-2,\ 3).$

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