Question

The function $ f(x) = x^3 - 3x $ is:

• A. increasing in \( (-\infty, -1) \cup (1, \infty) \) and decreasing in \( (-1, 1) \)
• B. decreasing in \( (-\infty, -1) \cup (1, \infty) \) and increasing in \( (-1, 1) \)
• C. increasing in \( (0, \infty) \) and decreasing in \( (-\infty, 0) \)
• D. decreasing in \( (0, \infty) \) and increasing in \( (-\infty, 0) \)

Hint:

To determine where a function is increasing or decreasing, find the critical points by setting the first derivative equal to zero and analyze the sign of the derivative.

Answer 1

The Correct Option is A

Step 1:
To find where the function is increasing or decreasing, we first compute the first derivative of the function \( f(x) = x^3 - 3x \): \[ f'(x) = 3x^2 - 3 \]
Step 2:
Set \( f'(x) = 0 \) to find the critical points: \[ 3x^2 - 3 = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1 \] Thus, the critical points are \( x = -1 \) and \( x = 1 \).
Step 3:
To determine whether the function is increasing or decreasing, we analyze the sign of \( f'(x) \) in each of the intervals defined by the critical points: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \). For \( x \in (-\infty, -1) \), choose a test point \( x = -2 \): \[ f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9>0 \] Therefore, the function is increasing in \( (-\infty, -1) \). For \( x \in (-1, 1) \), choose a test point \( x = 0 \): \[ f'(0) = 3(0)^2 - 3 = -3<0 \] Therefore, the function is decreasing in \( (-1, 1) \). For \( x \in (1, \infty) \), choose a test point \( x = 2 \): \[ f'(2) = 3(2)^2 - 3 = 12 - 3 = 9>0 \] Therefore, the function is increasing in \( (1, \infty) \).
Step 4:
From the analysis above, we conclude that the function is increasing in \( (-\infty, -1) \cup (1, \infty) \) and decreasing in \( (-1, 1) \).