Question

Find the intervals in which the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) is strictly increasing or decreasing?

Hint:

To determine where a function is increasing or decreasing, find the first derivative and use the sign of the derivative in different intervals.

Answer 1

Step 1: Find the first derivative of the function.
To determine the intervals where the function is increasing or decreasing, we first need to find the first derivative of \( f(x) \). The first derivative tells us the rate of change of the function. The given function is: \[ f(x) = 2x^3 - 15x^2 + 36x + 1 \] Differentiate it with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( 2x^3 - 15x^2 + 36x + 1 \right) \] Using standard differentiation rules: \[ f'(x) = 6x^2 - 30x + 36 \] Step 2: Solve for critical points.
To find the critical points, we set the first derivative equal to zero and solve for \( x \): \[ 6x^2 - 30x + 36 = 0 \] Divide through by 6 to simplify: \[ x^2 - 5x + 6 = 0 \] Now, solve this quadratic equation by factoring: \[ (x - 2)(x - 3) = 0 \] So, the critical points are \( x = 2 \) and \( x = 3 \). Step 3: Test intervals.
Now, we will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \). - For \( x \in (-\infty, 2) \), choose \( x = 1 \): \[ f'(1) = 6(1)^2 - 30(1) + 36 = 6 - 30 + 36 = 12 \quad (\text{positive, so increasing}) \] - For \( x \in (2, 3) \), choose \( x = 2.5 \): \[ f'(2.5) = 6(2.5)^2 - 30(2.5) + 36 = 6(6.25) - 75 + 36 = 37.5 - 75 + 36 = -1.5 \quad (\text{negative, so decreasing}) \] - For \( x \in (3, \infty) \), choose \( x = 4 \): \[ f'(4) = 6(4)^2 - 30(4) + 36 = 6(16) - 120 + 36 = 96 - 120 + 36 = 12 \quad (\text{positive, so increasing}) \] Step 4: Conclusion.
The function is:
- Increasing on \( (-\infty, 2) \) and \( (3, \infty) \)
- Decreasing on \( (2, 3) \)