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

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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.

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