Find all relative extrema of the function of the following functiona. 𝑓(𝑥) = 2𝑥3 − 3𝑥2 − 36𝑥 + 14 b

To find the relative extrema of the function f(x) = 2x³ - 3x² - 36x + 14, we first need to find the derivative of the function.

Step 1: Find the derivative of the function The derivative of the function f(x) = 2x³ - 3x² - 36x + 14 is f'(x) = 6x² - 6x - 36.

Step 2: Set the derivative equal to zero and solve for x Setting the derivative equal to zero gives us the critical points of the function.

6x² - 6x - 36 = 0 Divide the entire equation by 6 to simplify: x² - x - 6 = 0 This can be factored to: (x - 3)(x + 2) = 0 Setting each factor equal to zero gives the solutions x = 3 and x = -2.

Step 3: Use the second derivative test to classify each critical point as a local minimum, local maximum, or neither. The second derivative of the function is f''(x) = 12x - 6.

For x = 3, f''(3) = 12*3 - 6 = 30, which is greater than zero, so x = 3 is a local minimum.

For x = -2, f''(-2) = 12*(-2) - 6 = -30, which is less than zero, so x = -2 is a local maximum.

So, the function f(x) = 2x³ - 3x² - 36x + 14 has a local maximum at x = -2 and a local minimum at x = 3.