Question
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 2x^3 - 12x^2 + 18x + 15$
$f(x) = 2x^3 - 12x^2 + 18x + 15$
✓
Answer
$f(x) = 2x^3 - 12x^2 + 18x + 15$
$f'(x) = 6x^2 - 24x + 18$
$= 6(x^2 - 4x + 3)$
$= 6(x - 1)(x - 3)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$⇒ 6(x - 1)(x - 3) > 0$
$⇒ (x - 1)(x - 3) > 0$
$[$Since, $6 > 0, 6(x - 1)(x - 3) > 0 ⇒ (x - 1)(x - 3) > 0]$
$⇒ x < 1$ or $x > 3$
$\Rightarrow\text{x}\in(-\infty,0)\cup(3,\infty).$
So, $f(x)$ is increasing on $(-\infty,0)\cup(3,\infty).$
For $f(x)$ to be decreasing, we must have
$f'(x) < 0$
$⇒ 6(x - 1)(x - 3) < 0$
$⇒ (x - 1)(x - 3) < 0$
$[$Since, $6 > 0, 6(x - 1)(x - 3) < 0 ⇒ (x - 1)(x - 3) < 0]$
$⇒ x < 1$ or $x < 3$
$\Rightarrow\text{x}\in(1,3)$
So, $f(x)$ is decreasing on $(1, 3).$
$f'(x) = 6x^2 - 24x + 18$
$= 6(x^2 - 4x + 3)$
$= 6(x - 1)(x - 3)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$⇒ 6(x - 1)(x - 3) > 0$
$⇒ (x - 1)(x - 3) > 0$
$[$Since, $6 > 0, 6(x - 1)(x - 3) > 0 ⇒ (x - 1)(x - 3) > 0]$
$⇒ x < 1$ or $x > 3$
$\Rightarrow\text{x}\in(-\infty,0)\cup(3,\infty).$
So, $f(x)$ is increasing on $(-\infty,0)\cup(3,\infty).$
For $f(x)$ to be decreasing, we must have
$f'(x) < 0$
$⇒ 6(x - 1)(x - 3) < 0$
$⇒ (x - 1)(x - 3) < 0$
$[$Since, $6 > 0, 6(x - 1)(x - 3) < 0 ⇒ (x - 1)(x - 3) < 0]$
$⇒ x < 1$ or $x < 3$
$\Rightarrow\text{x}\in(1,3)$
So, $f(x)$ is decreasing on $(1, 3).$
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