Question

For the following function,

give the coordinates of any critical points and classify each point as a relative maximum, a relative minimum, or neither;
h(x)=2x³-6x

Answer 1

The critical points of the function h(x)=2x³-6x are (-1, -8) and (1, -4). The point (-1, -8) is a relative maximum and the point (1, -4) is a relative minimum.

Step-by-step explanation:

The critical points of the function h(x)=2x³-6x can be found by finding the values of x where the derivative of h(x) is equal to zero. So, let's find the derivative of h(x). We differentiate h(x) with respect to x to get h'(x)=6x²-6.

To find the critical points, we set h'(x) equal to zero and solve for x. So, 6x²-6=0. Solving this quadratic equation gives us x=±1.

Therefore, the coordinates of the critical points are (-1, -8) and (1, -4). To classify each point, we can use the second derivative test. The second derivative is h''(x)=12x. Since h''(-1)=-12 and h''(1)=12, we can conclude that the point (-1, -8) is a relative maximum and the point (1, -4) is a relative minimum.