Question

Find the x and y coordinates of this function's relative extrema.

f(x) = 2x^3 - 3x^2 - 36x + 1

Relative Minimum at (x, y)
Relative Maximum at (x, y)

Answer 1

To determine the relative extrema of the given function, the first derivative is calculated and set to zero to find critical points, then the second derivative test is used to classify each point as a minimum or maximum, followed by substituting these points in the original function to find y-coordinates.

Step-by-step explanation:

To find the x and y coordinates of the relative extrema of the function f(x) = 2x3 - 3x2 - 36x + 1, we need to first determine the first derivative of the function to find the critical points. The first derivative is found by applying the power rule to each term:

f'(x) = 6x2 - 6x - 36

We then set the first derivative equal to zero to find the critical points:

0 = 6x2 - 6x - 36

This can be factored or solved using the quadratic formula. After finding the critical points, we use the second derivative, f''(x) = 12x - 6, to test whether each critical point is a relative minimum or maximum.

A critical point x is a relative minimum if f''(x) is positive, and a relative maximum if f''(x) is negative. After finding the nature of each critical point, we substitute them back into the original function f(x) to find the corresponding y-coordinate.