Question
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 5x^3 - 15x^2 - 120x + 3.$
$f(x) = 5x^3 - 15x^2 - 120x + 3.$
✓
Answer
$f(x) = 5x^3 - 15x^2 - 120x + 3 $
$f'(x) = 15x^2 - 30x - 120 $
$= 15(x^2 - 2x - 8) $
$= 15(x - 4)(x + 2)$
For $f(x)$ to be increasing,
we must have $f'(x) > 0 $
$\Rightarrow 15(x - 4)(x + 2) > 0 $
$\Rightarrow (x - 4)(x + 2) > 0 [$Since, $15 > 0, 15(x - 4)(x + 2) > 0 \Rightarrow (x - 4)(x + 2) > 0]$
$\Rightarrow x < -2$ or $x > 4$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(4,\infty)$
So, $f(x)$ is increasing on $\text{x}\in(-\infty,-2)\cap(4,\infty).$
For $f(x)$ to be decreasing,
we must have, $f'(x) < 0 $
$\Rightarrow 6(x - 4)(x + 2) < 0 $
$\Rightarrow (x - 4)(x + 2) < 0 [$Since, $15 > 0, 15(x - 4)(x + 2) > 0 \Rightarrow (x - 4)(x + 2) > 0] $
$\Rightarrow -2 < x < 4$
$\Rightarrow\text{x}\in(-2,4)$
So, f(x) is decreasing on $\text{x}\in(-2,4).$
$f'(x) = 15x^2 - 30x - 120 $
$= 15(x^2 - 2x - 8) $
$= 15(x - 4)(x + 2)$
For $f(x)$ to be increasing,
we must have $f'(x) > 0 $
$\Rightarrow 15(x - 4)(x + 2) > 0 $
$\Rightarrow (x - 4)(x + 2) > 0 [$Since, $15 > 0, 15(x - 4)(x + 2) > 0 \Rightarrow (x - 4)(x + 2) > 0]$
$\Rightarrow x < -2$ or $x > 4$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(4,\infty)$
So, $f(x)$ is increasing on $\text{x}\in(-\infty,-2)\cap(4,\infty).$
For $f(x)$ to be decreasing,
we must have, $f'(x) < 0 $
$\Rightarrow 6(x - 4)(x + 2) < 0 $
$\Rightarrow (x - 4)(x + 2) < 0 [$Since, $15 > 0, 15(x - 4)(x + 2) > 0 \Rightarrow (x - 4)(x + 2) > 0] $
$\Rightarrow -2 < x < 4$
$\Rightarrow\text{x}\in(-2,4)$
So, f(x) is decreasing on $\text{x}\in(-2,4).$
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