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 - x^2/2 + 30x - 8.

The derivative of the function y, denoted as y', is found by applying the power rule of differentiation, which states that the derivative of x^n is n*x^(n-1).

So, the derivative of -x^3/3 is -3x^2/3 = -x^2.
The derivative of -x^2/2 is -2x/2 = -x.
The derivative of 30x is 30.
The derivative of -8 is 0, because the derivative of a constant is 0.

So, the derivative of the function y is y' = -x^2 - x + 30.

Next, we find the critical points by setting the derivative equal to zero and solving for x:

-x^2 - x + 30 = 0.

This is a quadratic equation, and we can solve it by using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a), where a, b, and c are the coefficients of the quadratic equation. In this case, a = -1, b = -1, and c = 30.

So, the solutions are x = [1 ± sqrt((1)^2 - 4*(-1)*30)] / (2*(-1)) = [1 ± sqrt(1 + 120)] / -2 = [1 ± sqrt(121)] / -2 = [1 ± 11] / -2.

So, the critical points are x = -6 and x = -5.

Finally, we use the first derivative test to classify each critical point. We pick a number less than -6, between -6 and -5, and greater than -5, and plug these numbers into the derivative to see if the function is increasing or decreasing.

If we pick x = -7, -x^2 - x + 30 = -49 - (-7) + 30 = -12, which is negative, so the function is decreasing.

If we pick x = -5.5, -x^2 - x + 30 = -30.25 - (-5.5) + 30 = 4.75, which is positive, so the function is increasing.

If we pick x = -4, -x^2 - x + 30 = -16 - (-4) + 30 = 18, which is positive, so the function is increasing.

So, at x = -6, the function changes from decreasing to increasing, so there is a relative minimum. At x = -5, the function does not change direction, so it is neither a relative minimum nor a relative maximum.

Question

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 - x^2/2 + 30x - 8.

The derivative of the function y, denoted as y', is found by applying the power rule of differentiation, which states that the derivative of x^n is n*x^(n-1).

So, the derivative of -x^3/3 is -3x^2/3 = -x^2.
The derivative of -x^2/2 is -2x/2 = -x.
The derivative of 30x is 30.
The derivative of -8 is 0, because the derivative of a constant is 0.

So, the derivative of the function y is y' = -x^2 - x + 30.

Next, we find the critical points by setting the derivative equal to zero and solving for x:

-x^2 - x + 30 = 0.

This is a quadratic equation, and we can solve it by using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a), where a, b, and c are the coefficients of the quadratic equation. In this case, a = -1, b = -1, and c = 30.

So, the solutions are x = [1 ± sqrt((1)^2 - 4*(-1)*30)] / (2*(-1)) = [1 ± sqrt(1 + 120)] / -2 = [1 ± sqrt(121)] / -2 = [1 ± 11] / -2.

So, the critical points are x = -6 and x = -5.

Finally, we use the first derivative test to classify each critical point. We pick a number less than -6, between -6 and -5, and greater than -5, and plug these numbers into the derivative to see if the function is increasing or decreasing.

If we pick x = -7, -x^2 - x + 30 = -49 - (-7) + 30 = -12, which is negative, so the function is decreasing.

If we pick x = -5.5, -x^2 - x + 30 = -30.25 - (-5.5) + 30 = 4.75, which is positive, so the function is increasing.

If we pick x = -4, -x^2 - x + 30 = -16 - (-4) + 30 = 18, which is positive, so the function is increasing.

So, at x = -6, the function changes from decreasing to increasing, so there is a relative minimum. At x = -5, the function does not change direction, so it is neither a relative minimum nor a relative maximum.

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