Analysis of the Function f(x) = 3x^4 - 4x^3 - 12x^2 + 5

To determine where the function f(x) = 3x^4 - 4x^3 - 12x^2 + 5 is increasing and decreasing, as well as finding its local maximum and minimum values, we need to analyze its first and second derivatives.

Step 1: Find the First Derivative
To find the first derivative, we differentiate the function f(x) with respect to x.

f'(x) = 12x^3 - 12x^2 - 24x

Step 2: Find the Critical Points
To find the critical points, we set the first derivative f'(x) equal to zero and solve for x.

12x^3 - 12x^2 - 24x = 0

Let's factor out the common factor of 12x:

12x(x^2 - x - 2) = 0

Now, we have two possibilities:
1. 12x = 0
This gives us x = 0 as a critical point.

2. x^2 - x - 2 = 0
We can factor this quadratic equation as (x - 2)(x + 1) = 0
This gives us x = 2 and x = -1 as additional critical points.

Therefore, the critical points are x = 0, x = 2, and x = -1.

Step 3: Determine the Intervals of Increase and Decrease
To determine the intervals of increase and decrease, we use the first derivative test. We choose test points in each interval and evaluate the sign of the derivative.

Interval (-∞, -1)
We choose a test point x = -2:
f'(-2) = 12(-2)^3 - 12(-2)^2 - 24(-2) = -96 + 48 + 48 = 0
Since f'(-2) = 0, the function is neither increasing nor decreasing in this interval.

Interval (-1, 0)
We choose a test point x = -0.5:
f'(-0.5) = 12(-0.5)^3 - 12(-0.5)^2 - 24(-0.5) = -1.5 + 0.75 + 12 = 11.25
Since f'(-0.5) > 0, the function is increasing in this interval.

Interval (0, 2)
We choose a test point x = 1:
f'(1) = 12(1)^3 - 12(1)^2 - 24(1) = 12 - 12 - 24 = -24
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Interval (2, ∞)
We choose a test point x = 3:
f'(3) = 12(3)^3 - 12(3)^2 - 24(3) = 108 - 108 - 72 = -72
Since f'(3) < 0,="" the="" function="" is="" decreasing="" in="" 0,="" the="" function="" is="" decreasing="" />