Find the max or min value for y given the restictions on x.

y= x^3 -x^2 -6x

a) 0<x<3 min

b) -2<x<0 max

what is tttt?

y = x^3 - x^2 - 6x = 0(x+2)(x-3)

y' = 3x^2 - 2x - 6

y' = 0 at approximately 1.7, -1.1 (1 +- sqrt(19))/3.

Knowing what we do about the shape of cubics, we can say that a local max is at x=-1.1, and a local min is at x=1.7

Note how the given intervals correspond to exactly one "hump" of the graph, so the local min/max is the min/max for that interval.

I don't understand the y' stuff. I'm only in algebra. Is there another way?

how do u convert 63 yards into ft

To find the maximum or minimum value for y given the restrictions on x, we can use calculus by finding the critical points and evaluating the endpoints of the given intervals.

a) For the interval 0 < x < 3, we first find the derivative of y with respect to x:

dy/dx = 3x^2 - 2x - 6

To find the critical point(s), we set the derivative equal to zero and solve for x:

0 = 3x^2 - 2x - 6

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

x = (-(-2) ± √((-2)^2 - 4(3)(-6))) / (2(3))
x = (2 ± √(4 + 72)) / 6
x = (2 ± √76) / 6
x = (1 ± √19) / 3

Since the interval is 0 < x < 3, we only need to consider the critical point(s) that fall within this interval. As 1 - √19/3 is negative and outside the interval, we discard it. Therefore, we have one critical point within the interval, x = (1 + √19) / 3.

Next, we evaluate the function y at the critical point and the endpoints of the interval:

y(0) = 0^3 - 0^2 - 6(0) = 0
y(3) = 3^3 - 3^2 - 6(3) = 0

Comparing the values, we can see that the maximum or minimum value for y within the given interval is 0. In this case, it is a minimum (since the signs change from negative to positive at the critical point).

b) For the interval -2 < x < 0, we follow the same steps:

1. Find the derivative of y:
dy/dx = 3x^2 - 2x - 6

2. Set the derivative equal to zero and solve for x:
0 = 3x^2 - 2x - 6

Using the quadratic formula again, we get:
x = (2 ± √(4 + 72)) / 6
x = (2 ± √76) / 6
x = (1 ± √19) / 3

Since the interval is -2 < x < 0, we only need to consider the critical point(s) that fall within this interval. As 1 + √19/3 is positive and outside the interval, we discard it. Therefore, we have one critical point within the interval, x = (1 - √19) / 3.

Next, we evaluate the function y at the critical point and the endpoints of the interval:

y(-2) = (-2)^3 - (-2)^2 - 6(-2) = -2
y(0) = 0^3 - 0^2 - 6(0) = 0

Comparing the values, we can see that the maximum or minimum value for y within the given interval is -2. In this case, it is a maximum (since the signs change from positive to negative at the critical point).

To summarize:
a) For the interval 0 < x < 3, the minimum value for y is 0.
b) For the interval -2 < x < 0, the maximum value for y is -2.