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For f(x)=2 x³-3 x²-36 x Determine the intervals of increase/decrease, and state the critical points

User D V Ramana
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Final answer:

To find the intervals of increase/decrease and critical points of the function f(x) = 2x³ - 3x² - 36x, find the derivative, solve for critical points, and analyze the sign changes.

Step-by-step explanation:

To determine the intervals of increase/decrease and critical points of the function f(x) = 2x³ - 3x² - 36x, we need to find the derivative of the function and analyze its sign changes.

1. Find the derivative of f(x) using the power rule:
f'(x) = 6x² - 6x - 36

2. Set f'(x) equal to zero and solve for x to find the critical points:
6x² - 6x - 36 = 0
Factoring out a common factor of 6: 6(x² - x - 6) = 0
Applying the zero product property:
x² - x - 6 = 0
Factoring the quadratic equation:
(x - 3)(x + 2) = 0
The critical points are x = 3 and x = -2.

3. Analyze the sign changes of f'(x) to determine the intervals of increase/decrease:
Use a sign chart or test different intervals to find that f'(x) is positive for x < -2, negative between -2 < x < 3, and positive for x > 3.

Therefore, the function f(x) = 2x³ - 3x² - 36x is increasing for x < -2 and x > 3, and decreasing between -2 < x < 3. The critical points are x = 3 and x = -2.

User Srd
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