The function f(x) is defined as f(x) = -2(x+2)(x-1)^2 = -2(x^3 -3x + 2) on the open interval (-3,3). 1. Determine the x-coordinate of the absolute minimum of f(x) in the open interval (-3,3). Justify your answer. 2. Find all values of x for which f(x) is concave down. Justify your answer. 3. Let g(x) be defined as g(x) = abs(f'(x)), the absolute value of the derivative of f(x), in the open interval (-3,3). Determine the (x,y) coordinate of the relative maxima of g(x) in the open interval. Justify your answer. 4. Find the x-values of the relative minima of g(x)? Justify your answer.

are you having trouble with this? It looks like straightforward use of derivatives to find critical values and concavity. What do you get?

To answer these questions, we will need to follow the step-by-step procedure. Let's solve each question one by one.

1. Determine the x-coordinate of the absolute minimum of f(x) in the open interval (-3,3). Justify your answer.

To find the absolute minimum, we need to locate critical points within the given interval and compare their function values.

Step 1: Find the critical points by setting the derivative of f(x) equal to zero.
f'(x) = -6x^2 + 12x - 6 = 0

Step 2: Solve for x by factoring or using the quadratic formula.
-6x^2 + 12x - 6 = 0
x^2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1

Step 3: Check the endpoints of the interval.
Checking x = -3:
f(-3) = -2((-3+2)(-3-1)^2) = -2(1 * 16) = -32

Checking x = 3:
f(3) = -2((3+2)(3-1)^2) = -2(5 * 4) = -40

Step 4: Compare the function values at critical points and endpoints.
f(1) = -2((1+2)(1-1)^2) = 0

From our calculations, the function value at x = 1 is 0, which is the absolute minimum within the interval (-3,3).

2. Find all values of x for which f(x) is concave down. Justify your answer.

We can determine the concavity of f(x) by analyzing the second derivative, f''(x).

Step 1: Find the second derivative of f(x).
f'(x) = -6x^2 + 12x - 6
f''(x) = -12x + 12

Step 2: Set f''(x) less than zero to find where f(x) is concave down.
-12x + 12 < 0
-12x < -12
x > 1

From our calculations, f(x) is concave down for x > 1.

3. Let g(x) be defined as g(x) = abs(f'(x)), the absolute value of the derivative of f(x), in the open interval (-3,3). Determine the (x,y) coordinate of the relative maxima of g(x) in the open interval. Justify your answer.

To find the relative maximas of g(x), we need to locate the critical points of g(x).

Step 1: Find the derivative of f(x).
f'(x) = -6x^2 + 12x - 6

Step 2: Find the critical points of g(x) by setting f'(x) equal to zero.
-6x^2 + 12x - 6 = 0

Step 3: Solve for x.
Divide the equation by -6:
x^2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1

Step 4: Find the corresponding y-coordinate.
g(1) = abs(f'(1)) = abs(-6(1)^2 + 12(1) - 6) = abs(0) = 0

So, the coordinate of the relative maxima of g(x) is (1,0).

4. Find the x-values of the relative minima of g(x)? Justify your answer.

Since the absolute value of a function cannot have a relative minimum, there are no x-values of relative minima for g(x).

To answer these questions, we'll go step by step using calculus. Let's begin with question 1:

1. To find the x-coordinate of the absolute minimum of f(x) in the open interval (-3,3), we need to find the critical points of f(x) and determine which one corresponds to the absolute minimum.

First, let's find the derivative of f(x):
f'(x) = -6(x^2 - 1)

Setting f'(x) = 0, we can find the critical points:
-6(x^2 - 1) = 0
x^2 - 1 = 0
x^2 = 1
x = ±1

Now, we need to check which critical point corresponds to the absolute minimum. We can do this by analyzing the sign changes of f'(x) around the critical points.

For x < -1, f'(x) is positive.
For -1 < x < 1, f'(x) is negative.
For x > 1, f'(x) is positive.

Hence, we have a local minimum at x = -1.

We also need to check the behavior at the endpoints of the interval (-3,3), which are x = -3 and x = 3.

For x = -3, f'(-3) = -6((-3)^2 -1) = -72 <0
For x = 3, f'(3) = -6((3)^2 -1) = -72 <0

Since f'(x) is negative at both endpoints and has a local minimum at x = -1, the relative minimum must be at either x = -1 or at the endpoints.

Now let's move on to question 2:

2. To find the values of x for which f(x) is concave down, we need to determine the points where the second derivative of f(x) is negative.

Taking the second derivative of f(x):
f''(x) = -12x

For f''(x) to be negative, we need x to be positive. Therefore, f(x) is concave down for x > 0.

Now let's tackle question 3:

3. The function g(x) is defined as the absolute value of the derivative of f(x), g(x) = |f'(x)|.

To find the relative maxima of g(x), we need to determine where g'(x) = 0 or where there is a change in the sign of g'(x).

Let's find g'(x):
g'(x) = f''(x) if f'(x) > 0
-f''(x) if f'(x) < 0

Since f'(x) is negative in the interval (-3,3), g'(x) = -f''(x) = 12x.

Setting g'(x) = 0:
12x = 0
x = 0

Now, let's analyze the sign changes of g'(x) around x = 0.

For x < 0, g'(x) is negative.
For x > 0, g'(x) is positive.

Hence, x = 0 is a relative minimum.

Finally, for question 4:

4. To find the x-values of the relative minima of g(x), we need to identify the critical points by checking where g''(x) = 0 or where there is a change in the sign of g''(x).

Taking the second derivative of g(x):
g''(x) = f'''(x) if f'(x) < 0
-f'''(x) if f'(x) > 0

In this case, since f'(x) is negative in the interval (-3,3), g''(x) = -f'''(x).

Therefore, g''(x) = 12.

Since g''(x) is a constant, it does not change sign. Thus, there are no relative minima of g(x) in the open interval (-3,3).

To summarize:
1. The x-coordinate of the absolute minimum of f(x) in the open interval (-3,3) is x = -1.
2. f(x) is concave down for x > 0.
3. The (x, y) coordinate of the relative maximum of g(x) in the open interval is (0, 12).
4. There are no relative minima of g(x) in the open interval (-3,3).