Finding Intervals Where a Function Decreases

To find the intervals where a function \(f(x)\) is decreasing, we need to determine where its first derivative, \(f'(x)\), is negative. A function \(f(x)\) decreases on an interval if \(f'(x) < 0\) for all \(x\) in that interval.

Step-by-Step Solution for Function Decreasing

The given function is \( f(x) = \frac{1}{12} (3x^4 + 4x^3 - 12x^2) \). First, let's find the derivative \(f'(x)\).

We use the power rule for differentiation: \(\frac{d}{dx}(x^n) = nx^{n-1}\).

\( f'(x) = \frac{d}{dx} \left( \frac{1}{12} (3x^4 + 4x^3 - 12x^2) \right) \)

\( f'(x) = \frac{1}{12} \frac{d}{dx} (3x^4 + 4x^3 - 12x^2) \)

\( f'(x) = \frac{1}{12} (3 \cdot 4x^{4-1} + 4 \cdot 3x^{3-1} - 12 \cdot 2x^{2-1}) \)

\( f'(x) = \frac{1}{12} (12x^3 + 12x^2 - 24x) \)

Now, we can simplify \(f'(x)\) by dividing each term inside the parenthesis by 12:

\( f'(x) = \frac{12x^3}{12} + \frac{12x^2}{12} - \frac{24x}{12} \)

\( f'(x) = x^3 + x^2 - 2x \)

Finding Critical Points for Function Analysis

Critical points are the points where the derivative \(f'(x)\) is equal to zero or undefined. For polynomial functions, the derivative is always defined. So, we set \(f'(x) = 0\) and solve for \(x\):

\( x^3 + x^2 - 2x = 0 \)

Factor out \(x\) from the expression:

\( x(x^2 + x - 2) = 0 \)

Now, factor the quadratic expression \(x^2 + x - 2\). We look for two numbers that multiply to -2 and add up to 1. These numbers are 2 and -1.

\( x(x+2)(x-1) = 0 \)

This equation is satisfied when any of the factors are zero. So, the critical points are:

• \(x = 0\)
• \(x + 2 = 0 \implies x = -2\)
• \(x - 1 = 0 \implies x = 1\)

The critical points are \(x = -2, x = 0, x = 1\).

Analyzing Intervals for Function Decrease

These critical points divide the number line into four intervals: \( (-\infty, -2) \), \( (-2, 0) \), \( (0, 1) \), and \( (1, \infty) \). We need to test the sign of \(f'(x)\) in each of these intervals to determine where the function is decreasing.

Conclusion: Function Decreasing Intervals

From the table, we see that \(f'(x)\) is negative in the intervals \( (-\infty, -2) \) and \( (0, 1) \). Therefore, the function \(f(x)\) is decreasing in these intervals.

The intervals where the function decreases are \( (-\infty, -2) \cup (0,1) \).

Revision Table: Key Concepts

Additional Information: Function Behavior and Extrema

Understanding where a function decreases is part of a larger analysis of function behavior using calculus. Besides decreasing intervals, we can also find increasing intervals (\(f'(x) > 0\)) and locate local extrema (local maximums and minimums) at critical points where the function changes from increasing to decreasing or vice versa.

• A local maximum occurs where the function changes from increasing to decreasing (\(f'(x)\) changes from positive to negative).
• A local minimum occurs where the function changes from decreasing to increasing (\(f'(x)\) changes from negative to positive).
• In this problem, the function changes from decreasing to increasing at \(x=-2\), suggesting a local minimum. It changes from increasing to decreasing at \(x=0\), suggesting a local maximum. It changes from decreasing to increasing at \(x=1\), suggesting another local minimum.

Analyzing the sign of the derivative \(f'(x)\) across intervals is a fundamental technique in curve sketching and understanding function properties.