Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives1 Mark
Question
Find the interval of the function that is strictly increasing or decreasing: $10 - 6x - 2x^2$
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Answer
It is given that function $f(x) = 10 - 6x - 2x^2$
$f\ ' (x) = -6 - 4x$
If $f\ ' (x) = 0,$ then we get,
$\Rightarrow x=\frac{-3}{2}$
So, the point $x=\frac{-3}{2}$ divides the real line into two disjoint intervals, $\left(-\infty, \frac{-3}{2}\right)$ and $\left(\frac{-3}{2}, \infty\right)$
So, in interval $\left(-\infty, \frac{-3}{2}\right)$
$x < \frac{-3}{2}$
$\Rightarrow -4x > 6$
$\Rightarrow-6 - 4x > 0$
i.e, $f\ ' (x) > 0$
Therefore, the given function $(f) $ is strictly increasing in interval $\left(-\infty, \frac{-3}{2}\right)$ and in interval $\left(\frac{-3}{2}, \infty\right)$
$f\ ' (x) = -6 - 4x < 0$
Therefore, the given function $(f)$ is strictly decreasing in interval $\left(\frac{-3}{2}, \infty\right)$
Thus, function is strictly increasing in $\left(-\infty, \frac{-3}{2}\right)$and strictly decreasing in $\left(\frac{-3}{2}, \infty\right)$
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