Question

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) (−[infinity],−8],[(7,[infinity])] Find the interval on which f is decreasing. (Enter your answer in interval notation.)

Answer 1

Answer:

The function f(x) = 2x³ + 3x² - 336x

is

Increasing on: (-∞, -8) U (7, ∞)

and

Decreasing on: (-8, 7)

Step-by-step explanation:

Given the function

f(x) = 2x³ + 3x² - 336x

We need to find the interval on which f is increasing, and the interval on which it is decreasing.

First, we take the derivative of f with respect to x to obtain

f'(x) = 6x² + 6x - 336

Next, set f'(x) = 0 and solve for x.

f'(x) = 6x² + 6x - 336 = 0

Dividing by 6, we have

x² + x - 56 = 0

(x - 7)(x + 8) = 0

x = 7 or x = -8

Now we have the interval

(-∞, -8) ∪ (-8, 7) ∪ (7, ∞)

Let us substitute a value from the interval (-∞, -8) into the derivative to determine if the function is increasing or decreasing.

Let us choose x = -9, then

f'(-9) = 6(-9)² + 6(-9) - 336

= 96 > 0

Since 96 is positive, the function is increasing on (-∞, -8).

Again, substitute a value from the interval (-8, 7) into the derivative to determine if the function is increasing or decreasing.

Choose -2

f'(-2) = 6(-2)² + 6(-2) - 336

= -324 < 0

Since -324 is negative, then the function is decreasing on (-8, 7)

Lastly, substitute a value from the interval (7, ∞) into the derivative

Choose x = 9

6(9)² + 6(9) - 336

= 204

The function is increasing on (8, ∞)

f'(x) > 0

Collating these, we have the function to be

Increasing on:

(-∞, -8) U (7, ∞)

Decreasing on:

(-8, 7)

Answer 2

Answer:

f is increasing = Solution Interval = x ∈ (-∞, -8] ∪ [7, ∞)

f is decreasing = Solution Interval = x ∈ [-8, 7]

Step-by-step explanation:

The given function is

f(x) = 2x³ + 3x² − 336x

We are required to find the interval on which f(x) is increasing and decreasing.

f(x) is increasing:

f(x) = 2x³ + 3x² − 336x

Taking the derivative of f(x) with respect to x yields,

f'(x) = 3*(2x²) + 2*(3x) - 336

f'(x) = 6x² + 6x) - 336

f'(x) ≥ 0

6x² + 6x - 336 ≥ 0

You can use either quadratic formula or use your calculator to solve the quadratic equation,

(x + 8) + (x - 7) ≥ 0

(x + 8) ≥ 0 or (x - 7) ≥ 0

x ≤ -8 or  x ≥ 7

so

x ∈ (-∞, -8]

x ∈ [7, ∞)

Solution Interval = x ∈ (-∞, -8] ∪ [7, ∞)

f(x) is decreasing:

f(x) = 2x³ + 3x² − 336x

We have already found the derivative f'(x)

f'(x) = 6x² + 6x - 336

f'(x) ≤ 0

6x² + 6x - 336 ≤ 0

(x - 7) + (x + 8) ≤ 0

(x - 7) ≤ 0 or (x + 8) ≤ 0

x ≤ 7 or x ≥ -8

-8 ≤ x ≤ 7

Solution Interval = x ∈ [-8, 7]