Question
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 8 + 36x + 3x^2 -2x^3$
$f(x) = 8 + 36x + 3x^2 -2x^3$
✓
Answer
$f(x) = 8 + 36x + 3x^2 -2x^3 $
$f'(x) = 36 + 6x - 6x^{2 }= -6(x^2 - x - 6) $
$= -6(x - 3)(x + 2)$ For $f(x)$ to be increasing,
we must have $f'(x) > 0 $
$\Rightarrow -6(x - 3)(x + 2) > 0 $
$\Rightarrow (x - 3)(x + 2) < 0 [$Since, $-6 > 0, -6(x - 3)(x + 2) > 0 \Rightarrow (x - 3)(x + 2) < 0]$
$\Rightarrow -2 < x < 3 \Rightarrow\text{x}\in(-2,3)$
So, $f(x)$ is increasing on $(-2, 3).$
For $f(x)$ to be decreasing,
we must have $f'(x) < 0 $
$\Rightarrow -6(x - 3)(x + 2) < 0$
$ \Rightarrow (x - 3)(x + 2) > 0$
$[$Since, $-6 < 0, -6(x - 3)(x + 2) < 0 \Rightarrow (x - 3)(x + 2) > 0] $
$\Rightarrow x < -2$ or $x > 3$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(3,\infty)$
So, $f(x)$ is decreasing on $(-\infty,-2)\cup(3,\infty).$
$f'(x) = 36 + 6x - 6x^{2 }= -6(x^2 - x - 6) $
$= -6(x - 3)(x + 2)$ For $f(x)$ to be increasing,
we must have $f'(x) > 0 $
$\Rightarrow -6(x - 3)(x + 2) > 0 $
$\Rightarrow (x - 3)(x + 2) < 0 [$Since, $-6 > 0, -6(x - 3)(x + 2) > 0 \Rightarrow (x - 3)(x + 2) < 0]$
$\Rightarrow -2 < x < 3 \Rightarrow\text{x}\in(-2,3)$
So, $f(x)$ is increasing on $(-2, 3).$
For $f(x)$ to be decreasing,
we must have $f'(x) < 0 $
$\Rightarrow -6(x - 3)(x + 2) < 0$
$ \Rightarrow (x - 3)(x + 2) > 0$
$[$Since, $-6 < 0, -6(x - 3)(x + 2) < 0 \Rightarrow (x - 3)(x + 2) > 0] $
$\Rightarrow x < -2$ or $x > 3$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(3,\infty)$
So, $f(x)$ is decreasing on $(-\infty,-2)\cup(3,\infty).$
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