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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES2 Marks
Question
Find the intervals in which the following function are strictly increasing or decreasing:
$-2x^3 - 9x^2 - 12x + 1$
✓

Answer

Given:$\text{f}\text{(x)} = -2\text{x}^3 -9\text{x}^2 - 12\text{x} + 1$ $\Rightarrow\ \text{f}'\text{(x)} = -6\text{x}^2 - 18\text{x} - 12$
$\Rightarrow\ \text{f}'\text{(x)} = -6(\text{x}^2 + 3\text{x} + 2) = -6(\text{x} + 1)(\text{x} + 2)\ \dots\text{(i)}$
$\text{Now } -6(\text{x} + 1)(\text{x} + 2) = 0\ \Rightarrow\ \text{x} = -1\text{ or } \text{x} = -2$
Therefore, we have three disjoint intervals $(-\infty,\ -2), (-2,\ -1)\text{ and }(-1, \ \infty).$
For interval $(-\infty,\ -2),$ from eq. $(i), f'(x), = (-) (-) (-) = (-) < 0$
Therefore, $f$ is strictly decreasing.
For interval $(-2, -1)$ from eq. $(i), f'(x) = (-) (-) (+) = (+) > 0$
Therefore, $f$ is strictly increasing
For interval $(-1, \ \infty) $ from eq. $(i), f'(x) = (-) (+) (+) = (-) < 0$
Therefore, $f$ is strictly decreasing.

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