Question

The cubic polynomial \( 2x^3 - 3x^2 - 36x + 28 \) is increasing in the range of \( x \). Find the interval where the function is increasing.

• A. \( x>2 \)
• B. \( x<-2 \)
• C. \( -2<x<2 \)
• D. \( x<0 \)

Hint:

To find the intervals where a function is increasing or decreasing, find the critical points by setting the first derivative equal to zero, and test the intervals around these points.

Answer 1

The Correct Option is C

We are given the cubic polynomial \( f(x) = 2x^3 - 3x^2 - 36x + 28 \), and we are asked to find the interval where the function is increasing.

1. Step 1: Find the derivative of the function. The first derivative of \( f(x) \) will tell us where the function is increasing or decreasing: \[ f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 36x + 28) \] Using standard differentiation rules: \[ f'(x) = 6x^2 - 6x - 36 \]

2. Step 2: Find the critical points by setting the derivative equal to zero. Set \( f'(x) = 0 \) to find the critical points: \[ 6x^2 - 6x - 36 = 0 \] Divide through by 6: \[ x^2 - x - 6 = 0 \] Factor the quadratic equation: \[ (x - 3)(x + 2) = 0 \] The critical points are \( x = 3 \) and \( x = -2 \).

3. Step 3: Test the intervals around the critical points. We now test the sign of \( f'(x) \) in the intervals \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \). - For \( x = -3 \) (in \( (-\infty, -2) \)), substitute into \( f'(x) \): \[ f'(-3) = 6(-3)^2 - 6(-3) - 36 = 54 + 18 - 36 = 36 \quad (\text{positive}) \] - For \( x = 0 \) (in \( (-2, 3) \)), substitute into \( f'(x) \): \[ f'(0) = 6(0)^2 - 6(0) - 36 = -36 \quad (\text{negative}) \] - For \( x = 4 \) (in \( (3, \infty) \)), substitute into \( f'(x) \): \[ f'(4) = 6(4)^2 - 6(4) - 36 = 96 - 24 - 36 = 36 \quad (\text{positive}) \]

4. Step 4: Determine the intervals of increase and decrease. - The function is increasing where \( f'(x)>0 \), which occurs in the intervals \( (-\infty, -2) \) and \( (3, \infty) \). - The function is decreasing in the interval \( (-2, 3) \). Thus, the function is increasing in the interval \( -2<x<2 \).