Find the 𝑥-values of both critical points of the following function. Then use the first derivative test to classify each of them as either a relative minimum, relative maximum, or neither.𝑦=−𝑥33−3𝑥22−2𝑥−7

To find the critical points of the function, we first need to find the derivative of the function. The function is y = -x^3/3 - 3x^2/2 - 2x - 7.

The derivative of the function is given by:

y' = -x^2 - 3x - 2

Setting the derivative equal to zero gives us the critical points:

0 = -x^2 - 3x - 2

Solving this quadratic equation will give us the x-values of the critical points. We can solve it by factoring, completing the square, or using the quadratic formula. The quadratic formula is the most straightforward method:

x = [-(-3) ± sqrt((-3)^2 - 4*(-1)(-2))]/(2(-1)) x = [3 ± sqrt(9 - 8)]/-2 x = [3 ± sqrt(1)]/-2 x = [3 ± 1]/-2 x = -2, -1

So the critical points are at x = -2 and x = -1.

To classify these points, we need to use the first derivative test. This involves checking the sign of the derivative on either side of each critical point.

For x = -2: Choose test points x = -3 and x = -1.5. y'(-3) = -(-3)^2 - 3*(-3) - 2 = -9 + 9 - 2 = -2, which is negative. y'(-1.5) = -(-1.5)^2 - 3*(-1.5) - 2 = -2.25 + 4.5 - 2 = 0.25, which is positive. Since the derivative changes from negative to positive, x = -2 is a relative minimum.

For x = -1: Choose test points x = -1.5 and x = -0.5. y'(-1.5) = -(-1.5)^2 - 3*(-1.5) - 2 = -2.25 + 4.5 - 2 = 0.25, which is positive. y'(-0.5) = -(-0.5)^2 - 3*(-0.5) - 2 = -0.25 + 1.5 - 2 = -0.75, which is negative. Since the derivative changes from positive to negative, x = -1 is a relative maximum.