How do you find decreasing intervals of

6x^3-63x^2-144x-2?

I found my increasing intervals from placing the derivatives i just can't seem to solve for the decrease

f decreasing means f' < 0

f' = 18x^2 - 126x - 144
= 18(x^2-7x-8)
= 18(x+1)(x-8)
Since that is a parabola which open upward, f' is negative between the roots.

f(x) is decreasing on the interval (-1,8)

This can be verified by looking at the graph:

http://www.wolframalpha.com/input/?i=6x^3-63x^2-144x-2

To find the decreasing intervals of a function, you need to examine the sign changes of its derivative. Here's how you can find the decreasing intervals for the function 6x^3 - 63x^2 - 144x - 2.

1. Find the derivative of the function:
Start by finding the derivative of the function using the power rule of differentiation. The derivative of 6x^3 - 63x^2 - 144x - 2 with respect to x is:
f'(x) = 18x^2 - 126x - 144.

2. Set the derivative equal to zero and solve for x:
To find the critical points, set the derivative f'(x) equal to zero and solve for x:
18x^2 - 126x - 144 = 0.

3. Solve the equation for x:
To solve this quadratic equation, you can factor it or use the quadratic formula. Factoring might be a bit challenging in this case, so it's recommended to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a.

Substituting the values a = 18, b = -126, and c = -144 into the quadratic formula, you get:
x = (-(-126) ± √((-126)^2 - 4 * 18 * (-144))) / (2 * 18).

Simplifying the equation yields the solutions:
x = (126 ± √(15876 + 10368)) / 36,
x = (126 ± √26244) / 36,
x = (126 ± 162) / 36.

Hence, x = (126 + 162) / 36 = 9, and x = (126 - 162) / 36 = -1.

4. Create an interval chart and analyze the intervals:
Create a number line and plot the critical points you found. These critical points divide the x-axis into intervals. Choose a test point from each interval and evaluate the derived function f'(x). Determine whether f'(x) is positive or negative in each interval:

Number line:
...(-∞)---(-1)---(9)---(+∞)...

Choose a test point from the interval (-∞, -1), such as x = -2:
Evaluate f'(-2):
f'(-2) = 18(-2)^2 - 126(-2) - 144,
f'(-2) = 72 + 252 - 144,
f'(-2) = 180.

Since f'(-2) = 180 is positive, the derivative is positive in the (-∞, -1) interval.

Choose a test point from the interval (-1, 9), such as x = 0:
Evaluate f'(0):
f'(0) = 18(0)^2 - 126(0) - 144,
f'(0) = 0 - 0 - 144,
f'(0) = -144.

Since f'(0) = -144 is negative, the derivative is negative in the (-1, 9) interval.

Choose a test point from the interval (9, +∞), such as x = 10:
Evaluate f'(10):
f'(10) = 18(10)^2 - 126(10) - 144,
f'(10) = 1800 - 1260 - 144,
f'(10) = 396.

Since f'(10) = 396 is positive, the derivative is positive in the (9, +∞) interval.

Therefore, the decreasing intervals for the function 6x^3 - 63x^2 - 144x - 2 are (-1, 9).