Question
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 - 12x2 + 18x + 15
f(x) = 2x3 - 12x2 + 18x + 15
✓
Answer
f(x) = 2x3 - 12x2 + 18x + 15
f'(x) = 6x2 - 24x + 18
= 6(x2 - 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) = 6x2 - 24x + 18
= 6(x2 - 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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