# 𝑓(𝑥)=2𝑥^3−3𝑥^2−12𝑥+2?

is decreasing on the interval ( , ).

It is increasing on the interval ( −∞, ) and the interval ( , ∞ ).

The function has a local maximum at

.

### 3 Answers

- llafferLv 74 weeks agoFavorite Answer
f(x) = 2x³ - 3x² - 12x + 2

To find the local minimums and maximums we can solve for the zeroes of the first derivative:

f'(x) = 6x² - 6x - 12

0 = 6x² - 6x - 12

Simplify by dividing both sides by 6:

0 = x² - x - 2

This factors:

0 = (x - 2)(x + 1)

x = -1 and 2

You can then check each of these x's to find the y's. The higher of the two will be your local maximum:

f(x) = 2x³ - 3x² - 12x + 2

f(-1) = 2(-1)³ - 3(-1)² - 12(-1) + 2 and f(2) = 2(2)³ - 3(2)² - 12(2) + 2

f(-1) = 2(-1) - 3(1) + 12 + 2 and f(2) = 2(8) - 3(4) - 24 + 2

f(-1) = -2 - 3 + 12 + 2 and f(2) = 16 - 12 - 24 + 2

f(-1) = 9 and f(2) = -18

The local minimum is (2, -18) and the local maximum is (-1, 9).

To find the intervals that are increasing or decreasing we look for the first derivative being greater than zero and less than zero.

We already have the equation for that and know the roots and know the factors so we can use the roots as pivots to check the ranges around it.

if x > 2 we have positive times positive which is positive

if -1 < x < 2 we have positive times negative which is negative

if x < -1 we have negative times negative which is positive

So the function is increasing on the interval of (-∞, -1) U (2, ∞)

The function is decreasing on the interval of (-1, 2)

- King LeoLv 74 weeks ago
f(x) = 2x³ - 3x² - 12x + 2

———> is decreasing on the interval where f’(x) < 0;f’(x) = 6x² - 6x - 12 6x² - 6x - 12 < 06 ( x - 2 )( x + 1 ) < 0-1 < x < 2is decreasing on the interval ( -1, 2 ).———> It is increasing on the interval where f’(x) > 0;;f’(x) = 6x² - 6x - 126x² - 6x - 12 > 06 ( x - 2 )( x + 1 ) > 0x < -1 and x > 2is increasing on the interval ( -∞, -1 ) and ( 2, ∞ )———> local maximum where f’(x) = 0 and f’’(x) < 0f’(x) = 6x² - 6x - 12 6x² - 6x - 12 = 06 ( x - 2 )( x + 1 ) = 0x = -1 or x = 2f’’(x) = 12x - 6f’’(-1) = -18f’’(2) = 18f(x) = 2x³ - 3x² - 12x + 2f(-1) = 9The function has a local maximum at ( -1, 9 )