For the function f(x) = 20x^3 − 3x^5, determine the interval(s) for which f(x) increases.
(–2, 2)
(–∞, –2) U (2, ∞)
(–2, 0) U (2, ∞)
(–∞, –2) U (0, 2)
To determine the interval(s) for which the function f(x) = 20x^3 − 3x^5 increases, we need to find the intervals where the derivative of the function is positive.
First, we find the derivative f'(x) of the function f(x):
f'(x) = (d/dx)(20x^3 − 3x^5)
f'(x) = 60x^2 - 15x^4
Next, we need to find the critical points of the function where f'(x) = 0 or is undefined. To do this, we set f'(x) = 0 and solve for x:
60x^2 - 15x^4 = 0
15x^2(4 - x^2) = 0
This equation is satisfied when x = 0 or x = ±2.
We can now create an interval table to determine the sign of f'(x) in each interval:
Interval (-∞, -2):
Choose a value less than -2, e.g., x = -3
Substitute the value into f'(x): f'(-3) = 60(-3)^2 - 15(-3)^4 = 540 - 405 = 135 (positive)
Therefore, f'(x) > 0 in the interval (-∞, -2).
Interval (-2, 0):
Choose a value between -2 and 0, e.g., x = -1
Substitute the value into f'(x): f'(-1) = 60(-1)^2 - 15(-1)^4 = 60 - 15 = 45 (positive)
Therefore, f'(x) > 0 in the interval (-2, 0).
Interval (0, 2):
Choose a value between 0 and 2, e.g., x = 1
Substitute the value into f'(x): f'(1) = 60(1)^2 - 15(1)^4 = 60 - 15 = 45 (positive)
Therefore, f'(x) > 0 in the interval (0, 2).
Interval (2, ∞):
Choose a value greater than 2, e.g., x = 3
Substitute the value into f'(x): f'(3) = 60(3)^2 - 15(3)^4 = 540 - 405 = 135 (positive)
Therefore, f'(x) > 0 in the interval (2, ∞).
Based on the sign of f'(x) in each interval, we observe that f(x) increases in the following intervals:
(–∞, –2) U (0, 2) U (2, ∞)
Therefore, the correct answer is:
(–∞, –2) U (0, 2) U (2, ∞)