If y = x3 - x2 - 7x - 3, what is the maximum value of y on the interval [-3, 3] ?

To find the maximum value of y on the interval [-3, 3], we need to follow these steps:

1. First, find the derivative of the function y = x^3 - x^2 - 7x - 3. The derivative, y', is 3x^2 - 2x - 7.

2. Set the derivative equal to zero and solve for x to find the critical points. So, 3x^2 - 2x - 7 = 0. Solving this quadratic equation can be a bit tricky, but you can use the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = 3, b = -2, and c = -7. Solving gives us two solutions, but only one, x = 1.1547, falls within our interval [-3, 3].

3. Evaluate the function at the critical point and at the endpoints of the interval. So, we find the values of y at x = -3, x = 1.1547, and x = 3.

4. The maximum value of y on the interval [-3, 3] is the largest of these three values.

Let's calculate these values:

• For x = -3, y = (-3)^3 - (-3)^2 - 7*(-3) - 3 = -27 - 9 + 21 - 3 = -18.
• For x = 1.1547, y = (1.1547)^3 - (1.1547)^2 - 7*1.1547 - 3 = 1.1547 - 1.3341 - 8.0829 - 3 = -11.2623.
• For x = 3, y = (3)^3 - (3)^2 - 7*3 - 3 = 27 - 9 - 21 - 3 = -6.

So, the maximum value of y on the interval [-3, 3] is -6.