Determining Intervals of Increasing and Decreasing | An Analysis of Derivatives for Math Function Behavior

Determine the intervals where f(x) is increasing or decreasing.Locate any local max or min values.

To determine the intervals where a function is increasing or decreasing, we need to analyze the behavior of the derivative of the function

To determine the intervals where a function is increasing or decreasing, we need to analyze the behavior of the derivative of the function. If the derivative is positive, the function is increasing on that interval; if it is negative, the function is decreasing.

Here are the steps to determine the intervals where f(x) is increasing or decreasing:

1. Differentiate the function f(x) to find its derivative, f'(x).

2. Set f'(x) equal to 0 and solve for x to find the critical points. These are the values of x where f'(x) changes sign or is undefined.

3. Choose test points within each interval created by the critical points and evaluate f'(x) at these points to determine if the derivative is positive or negative.

4. Create a sign chart based on the intervals determined in step 3. Mark the intervals where f'(x) is positive (+) or negative (-).

5. Based on the sign chart, determine the intervals where f(x) is increasing or decreasing.

6. To locate local max or min values, examine the behavior of f'(x) around the critical points. If f'(x) changes sign from positive to negative at a critical point, it indicates a local max value. If it changes sign from negative to positive, it indicates a local min value.

Let’s illustrate this process with an example:

Example:
Consider the function f(x) = x^3 – 3x^2 + 2x + 1.

1. Differentiate the function to find f'(x):
f'(x) = 3x^2 – 6x + 2.

2. Set f'(x) equal to 0 and solve for x to find the critical points:
3x^2 – 6x + 2 = 0.
Using the quadratic formula, we find:
x = (6 ± √(36 – 4(3)(2))) / (2(3)).
x = (6 ± √(36 – 24)) / 6.
x = (6 ± √12) / 6.
Simplifying, we get:
x = (6 ± 2√3) / 6.
Therefore, we have two critical points: x = 1 ± √3 / 3.

3. Choose test points within each interval:
Since we have two critical points, we can choose a test point to the left and right of each critical point. Let’s choose x = 0, x = 1, and x = 2 as our test points.

4. Evaluate f'(x) at the test points:
For x < 1 - √3 / 3, let's choose x = 0: f'(0) = 3(0)^2 - 6(0) + 2 = 2. Since f'(0) > 0, the interval (-∞, 1 – √3 / 3) is where f(x) is increasing.

For 1 – √3 / 3 < x < 1 + √3 / 3, let's choose x = 1: f'(1) = 3(1)^2 - 6(1) + 2 = -1. Since f'(1) < 0, the interval (1 - √3 / 3, 1 + √3 / 3) is where f(x) is decreasing. For x > 1 + √3 / 3, let’s choose x = 2:
f'(2) = 3(2)^2 – 6(2) + 2 = 2.
Since f'(2) > 0, the interval (1 + √3 / 3, ∞) is where f(x) is increasing.

Based on this information, we create a sign chart:
(+) (-) (+)
————–|———-|———-|———-

From the sign chart, we can observe the following:

– The function is increasing on the intervals (-∞, 1 – √3 / 3) and (1 + √3 / 3, ∞).
– The function is decreasing on the interval (1 – √3 / 3, 1 + √3 / 3).

5. The local max and min values can be determined based on the critical points:

The value of f(x) at x = 1 – √3 / 3 is a local maxima.
The value of f(x) at x = 1 + √3 / 3 is a local minima.

Therefore, the intervals of increasing and decreasing are:

Increasing intervals: (-∞, 1 – √3 / 3) and (1 + √3 / 3, ∞).
Decreasing interval: (1 – √3 / 3, 1 + √3 / 3).

Local max value: f(1 – √3 / 3).
Local min value: f(1 + √3 / 3).

These intervals and local extrema provide a complete description of the behavior of the function f(x).

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