Respuesta :
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:
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]
