Determine the intervals where the function f(x) = x^3 - 3x is increasing.
Practice Questions
1 question
Q1
Determine the intervals where the function f(x) = x^3 - 3x is increasing.
(-∞, -1)
(-1, 1)
(1, ∞)
(-∞, 1)
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the intervals where the function f(x) = x^3 - 3x is increasing.
Solution: f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).
Steps: 10
Step 1: Start with the function f(x) = x^3 - 3x.
Step 2: Find the derivative of the function, which tells us the rate of change. The derivative is f'(x) = 3x^2 - 3.
Step 3: Set the derivative equal to zero to find critical points: 3x^2 - 3 = 0.
Step 4: Solve for x. This gives us x^2 = 1, so x = ±1 (which means x = 1 and x = -1).
Step 5: Determine the intervals to test for increasing or decreasing. The critical points divide the number line into intervals: (-∞, -1), (-1, 1), and (1, ∞).
Step 6: Choose a test point from each interval to see if f'(x) is positive (increasing) or negative (decreasing).
Step 7: For the interval (-∞, -1), choose x = -2: f'(-2) = 3(-2)^2 - 3 = 9 > 0 (increasing).
Step 8: For the interval (-1, 1), choose x = 0: f'(0) = 3(0)^2 - 3 = -3 < 0 (decreasing).
Step 9: For the interval (1, ∞), choose x = 2: f'(2) = 3(2)^2 - 3 = 9 > 0 (increasing).
Step 10: Conclude that f(x) is increasing on the intervals (-∞, -1) and (1, ∞).
