Find the x-coordinate of the absolute minimum of f(x) = x^3 - x^2 - x +1 on the interval [-2, 2]

The function f(x) = x^3 - 2x^2 is increasing on which of the following intervals?
a)x<0 only
b)x>4/3 only
0<x<4/3
d)x<0 or x>4/3
e)x<0 or x>3/4

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To find the x-coordinate of the absolute minimum of the function f(x) = x^3 - x^2 - x + 1 on the interval [-2, 2], we can follow these steps:

1. Begin by finding the critical points of the function. Critical points occur where the derivative of the function is either zero or undefined.

2. Calculate the derivative of f(x) by taking the derivative of each term separately. The derivative of x^n is nx^(n-1).

f'(x) = 3x^2 - 2x - 1

3. Set the derivative equal to zero and solve for x to find the critical points:

3x^2 - 2x - 1 = 0

4. Factor or use the quadratic formula to find the solutions for x. In this case, factoring is not possible, so we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In the quadratic equation 3x^2 - 2x - 1 = 0, a = 3, b = -2, and c = -1.

Plugging in these values, we can solve for x:

x = (-(-2) ± √((-2)^2 - 4 * 3 * -1)) / (2 * 3)
= (2 ± √(4 + 12)) / 6
= (2 ± √16) / 6
= (2 ± 4) / 6

The two solutions for x are:
x = (2 + 4) / 6 = 6 / 6 = 1
x = (2 - 4) / 6 = -2 / 6 = -1/3

5. Now we need to check whether the critical points lie within the given interval [-2, 2]. Both -1/3 and 1 fall within the interval, so we consider them as potential minimum points.

6. Evaluate the function f(x) at the critical points and the endpoints of the interval [-2, 2].

f(-2) = (-2)^3 - (-2)^2 - (-2) + 1 = -1
f(-1/3) = (-1/3)^3 - (-1/3)^2 - (-1/3) + 1 = 14/27
f(1) = (1)^3 - (1)^2 - 1 + 1 = 0
f(2) = (2)^3 - (2)^2 - 2 + 1 = 1

7. Compare the values obtained in step 6 to find the minimum.

The lowest value is f(-2) = -1, which occurs at x = -2.

Therefore, the x-coordinate of the absolute minimum of f(x) = x^3 - x^2 - x + 1 on the interval [-2, 2] is x = -2.

For the second question regarding the function f(x) = x^3 - 2x^2, we need to determine the intervals on which the function is increasing. To find this, we can analyze the sign of the derivative of f(x).

1. Calculate the derivative of f(x) by taking the derivative of each term:

f'(x) = 3x^2 - 4x

2. Set the derivative equal to zero and solve for x to find the critical points:

3x^2 - 4x = 0

Factoring out x, we get:
x(3x - 4) = 0

The critical points are x = 0 and x = 4/3.

3. Now we have the critical points and we can analyze the sign of the derivative on each of the intervals:

- If x < 0, we can check a value such as x = -1 (which is less than 0):
f'(-1) = 3(-1)^2 - 4(-1) = 7

Since the value is positive, we can conclude that the function is increasing on the interval x < 0.

- If 0 < x < 4/3, we can check a value such as x = 1 (which is in the interval 0 < x < 4/3):
f'(1) = 3(1)^2 - 4(1) = -1

Since the value is negative, we can conclude that the function is decreasing on the interval 0 < x < 4/3.

- If x > 4/3, we can check a value such as x = 2 (which is greater than 4/3):
f'(2) = 3(2)^2 - 4(2) = 4

Since the value is positive, we can conclude that the function is increasing on the interval x > 4/3.

Based on these calculations, the function f(x) = x^3 - 2x^2 is increasing on the interval x < 0 only (option a).