7.

f(x)= 6x^3- 9x^2- 360x - 5

What interval is it decreasing?
What 2 intervals is it increasing?
What is the local max?

To determine the intervals in which a function is increasing or decreasing, we need to analyze the sign of its derivative. The derivative of a function gives us information about its slope or rate of change at any given point. By finding the critical points of the function, we can determine the intervals where it is increasing or decreasing.

Let's start by finding the derivative of the given function.

f(x) = 6x^3 - 9x^2 - 360x - 5

To get the derivative, we apply the power rule to each term:

f'(x) = 3 * 6x^2 - 2 * 9x - 360

Now we can find the critical points of the function by setting the derivative equal to zero and solving for x:

3 * 6x^2 - 2 * 9x - 360 = 0

18x^2 - 18x - 360 = 0

Simplifying the equation, we divide all terms by 18:

x^2 - x - 20 = 0

Now we can factor the quadratic equation:

(x - 5)(x + 4) = 0

Solving for x, we find two critical points:

x = 5 and x = -4

To determine the intervals of increasing and decreasing, we can create a sign chart. We evaluate the sign of the derivative in three separate regions:

1. For x < -4:
Pick a value to the left of -4, e.g., x = -5.
Evaluate f'(-5):
f'(-5) = 3 * 6(-5)^2 - 2 * 9(-5) - 360 = 420 - (-90) - 360 = 420 + 90 - 360 = 150

Since f'(-5) = 150 > 0, the derivative is positive in this region, indicating the function is increasing.

2. For -4 < x < 5:
Pick a value between -4 and 5, e.g., x = 0.
Evaluate f'(0):
f'(0) = 3 * 6(0)^2 - 2 * 9(0) - 360 = 0 - 0 - 360 = -360

Since f'(0) = -360 < 0, the derivative is negative in this region, indicating the function is decreasing.

3. For x > 5:
Pick a value to the right of 5, e.g., x = 6.
Evaluate f'(6):
f'(6) = 3 * 6(6)^2 - 2 * 9(6) - 360 = 648 - 324 - 360 = -36

Since f'(6) = -36 < 0, the derivative is negative in this region as well, indicating the function is decreasing.

Therefore, the function f(x) = 6x^3 - 9x^2 - 360x - 5 is decreasing in the interval (-4, 5) and increasing in the intervals (-∞, -4) and (5, ∞).

To find the local maximum, we need to look for critical points that are turning points. We can do this by examining the concavity of the function, which is determined by the second derivative.

Taking the derivative of the derivative, we obtain:

f''(x) = 2(18x - 9)

Setting this equal to zero and solving for x:

2(18x - 9) = 0
36x - 18 = 0
36x = 18
x = 1/2

Now, we need to determine the concavity of the function in different regions:

1. For x < 1/2:
Pick a value less than 1/2, e.g., x = 0.
Evaluate f''(0):
f''(0) = 2(18(0) - 9) = -18

Since f''(0) = -18 < 0, the function is concave down in this region.

2. For x > 1/2:
Pick a value greater than 1/2, e.g., x = 1.
Evaluate f''(1):
f''(1) = 2(18(1) - 9) = 18

Since f''(1) = 18 > 0, the function is concave up in this region.

From this analysis, we can conclude that the function has a local maximum at x = 1/2.

In summary:
- The interval in which the function is decreasing is (-4, 5).
- The intervals in which the function is increasing are (-∞, -4) and (5, ∞).
- The local maximum of the function occurs at x = 1/2.