9.2k views
2 votes
Given f(x)=2 x³+3 x²-12 x+4 on [-2,2] (a) The stationary points are

1 Answer

5 votes

Final answer:

To find the stationary points of the function f(x) = 2x^3 + 3x^2 - 12x + 4, we differentiate the function and solve for x when the derivative equals 0.

Step-by-step explanation:

The given function is f(x) = 2x^3 + 3x^2 - 12x + 4. To find the stationary points, we need to find the values of x where the derivative of the function equals 0.

To find the derivative of f(x), we differentiate each term and set it equal to 0:

f'(x) = 6x^2 + 6x - 12 = 0

Using the quadratic formula, we can solve for x:

x = (-b ± √(b^2 - 4ac))/(2a)

Substituting the values a = 6, b = 6, and c = -12, we get x = -2 and x = 1.

Therefore, the stationary points are x = -2 and x = 1.

User Ferhan
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories