Consider the following function.h(x) = x3 − 4x2 − 6x(a) Write the derivative formula.h'(x) = 3x2−8x−6 (b) Locate any relative extreme points. (Round your answers to three decimal places. If an answer does not exist, enter DNE.)relative maximum     (x, y) =

To find the relative extreme points, we first need to find the critical points of the function. Critical points occur where the derivative of the function is zero or undefined.

The derivative of the function h(x) = x^3 - 4x^2 - 6x is h'(x) = 3x^2 - 8x - 6.

Setting this equal to zero gives us the equation 3x^2 - 8x - 6 = 0.

We can solve this quadratic equation for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 3, b = -8, and c = -6.

Substituting these values into the quadratic formula gives us:

x = [8 ± sqrt((-8)^2 - 43(-6))] / (2*3) x = [8 ± sqrt(64 + 72)] / 6 x = [8 ± sqrt(136)] / 6 x = [8 ± 11.66] / 6

This gives us two solutions:

x = (8 + 11.66) / 6 = 3.277 x = (8 - 11.66) / 6 = -0.611

These are the x-values of the critical points.

To find the y-values, we substitute these x-values back into the original function h(x):

h(3.277) = (3.277)^3 - 4*(3.277)^2 - 63.277 = -13.849 h(-0.611) = (-0.611)^3 - 4(-0.611)^2 - 6*(-0.611) = 1.853

So the critical points are (3.277, -13.849) and (-0.611, 1.853).

To determine whether these are relative maximum or minimum points, we can use the second derivative test. The second derivative of h(x) is h''(x) = 6x - 8.

Substituting the x-values of the critical points gives us:

h''(3.277) = 63.277 - 8 = 11.662 h''(-0.611) = 6(-0.611) - 8 = -11.666

Since h''(3.277) > 0, the point (3.277, -13.849) is a relative minimum. Since h''(-0.611) < 0, the point (-0.611, 1.853) is a relative maximum.

So the relative maximum is (-0.611, 1.853) and the relative minimum is (3.277, -13.849).