How To Write Intervals Of Increase And Decrease

Diving into the world of calculus, understanding intervals of increase and decrease is a fundamental skill. These intervals tell us where a function is climbing uphill (increasing) or sliding downhill (decreasing). Mastering this concept unlocks deeper insights into a function's behavior and graphical representation, useful in optimization problems, curve sketching, and real-world applications.

Unveiling the Language of Functions: Increase, Decrease, and Constant

Before we plunge into the mechanics of identifying these intervals, let's clarify the vocabulary we'll be using.

• Increasing Function: A function f(x) is increasing on an interval if, for any two points x1 and x2 in the interval where x1 < x2, then f(x1) < f(x2). Think of it as moving from left to right on the graph; the y-values are getting larger.

• Decreasing Function: Conversely, a function f(x) is decreasing on an interval if, for any two points x1 and x2 in the interval where x1 < x2, then f(x1) > f(x2). As you move from left to right, the y-values are getting smaller.

• Constant Function: A function f(x) is constant on an interval if, for any two points x1 and x2 in the interval, f(x1) = f(x2). The graph is a horizontal line.

The Derivative: Your Key to Unlocking Function Behavior

The cornerstone of finding intervals of increase and decrease is the derivative of the function, denoted as f'(x). The derivative provides the slope of the tangent line to the function at any given point. This slope tells us the instantaneous rate of change of the function. Here's the crucial connection:

• If f'(x) > 0 on an interval, then f(x) is increasing on that interval. (Positive slope = uphill climb)
• If f'(x) < 0 on an interval, then f(x) is decreasing on that interval. (Negative slope = downhill slide)
• If f'(x) = 0 on an interval, then f(x) is constant on that interval. (Zero slope = horizontal line)

Step-by-Step Guide: Finding Intervals of Increase and Decrease

Now, let's break down the process into a series of manageable steps:

Step 1: Find the Derivative

This is the foundation. Calculate the derivative f'(x) of your function f(x). You'll need to apply the rules of differentiation (power rule, product rule, quotient rule, chain rule, etc.) depending on the complexity of the function.

Example:

Let's say our function is f(x) = x³ - 3x² - 9x + 5. Then, using the power rule, the derivative is f'(x) = 3x² - 6x - 9.

Step 2: Find Critical Points

Critical points are the x-values where the derivative is either equal to zero f'(x) = 0 or undefined. These points are crucial because they represent potential turning points of the function – where it might switch from increasing to decreasing, or vice versa.

• Set the derivative equal to zero and solve for x: This will give you the critical points where the tangent line is horizontal (slope = 0).
• Identify points where the derivative is undefined: This typically occurs when the derivative involves a fraction and the denominator is zero, or when dealing with functions like radicals or logarithms where the domain is restricted.

Example (Continuing from above):

To find the critical points, we set f'(x) = 0:

• 3x² - 6x - 9 = 0
• Divide by 3: x² - 2x - 3 = 0
• Factor: (x - 3)(x + 1) = 0
• Solve for x: x = 3 and x = -1

Therefore, our critical points are x = 3 and x = -1.

Step 3: Create a Sign Chart (or Number Line)

A sign chart is a visual tool to organize the information about the derivative's sign in different intervals.

• Draw a number line.
• Mark all critical points on the number line. These points divide the number line into intervals.
• Choose a test value within each interval (a number that's not a critical point).
• Evaluate the derivative f'(x) at each test value. The sign of f'(x) at the test value tells you the sign of f'(x) throughout the entire interval.
• Write the sign (+ or -) of f'(x) above each interval on the number line.

Example (Continuing from above):

Our critical points are x = -1 and x = 3. This divides the number line into three intervals:

• Interval 1: (-∞, -1)
• Interval 2: (-1, 3)
• Interval 3: (3, ∞)

Let's choose test values:

• Interval 1: Let's pick x = -2. f'(-2) = 3(-2)² - 6(-2) - 9 = 12 + 12 - 9 = 15. Since f'(-2) > 0, the derivative is positive in this interval.
• Interval 2: Let's pick x = 0. f'(0) = 3(0)² - 6(0) - 9 = -9. Since f'(0) < 0, the derivative is negative in this interval.
• Interval 3: Let's pick x = 4. f'(4) = 3(4)² - 6(4) - 9 = 48 - 24 - 9 = 15. Since f'(4) > 0, the derivative is positive in this interval.

Our sign chart looks like this:

        (-∞, -1)    (-1, 3)    (3, ∞)
---------------------------------------
f'(x):      +          -          +

Step 4: Determine Intervals of Increase and Decrease

Now, use the sign chart to determine the intervals where the function is increasing and decreasing.

• Increasing: If f'(x) > 0 on an interval, the function is increasing on that interval.
• Decreasing: If f'(x) < 0 on an interval, the function is decreasing on that interval.
• Constant: If f'(x) = 0 on an interval (or at a single point), the function is constant at that point. Note: A function is rarely constant on a true interval; it's more common to say it has a horizontal tangent at a specific point.

Example (Continuing from above):

Based on our sign chart:

• f(x) is increasing on the interval (-∞, -1) because f'(x) > 0.
• f(x) is decreasing on the interval (-1, 3) because f'(x) < 0.
• f(x) is increasing on the interval (3, ∞) because f'(x) > 0.

Step 5: Identify Local Maxima and Minima (Optional, but Often Useful)

Critical points can correspond to local maxima (peaks) or local minima (valleys) of the function.

• First Derivative Test: Examine the sign change of the derivative around each critical point.
  • If f'(x) changes from positive to negative at x = c, then f(x) has a local maximum at x = c.
  • If f'(x) changes from negative to positive at x = c, then f(x) has a local minimum at x = c.
  • If f'(x) does not change sign at x = c, then x = c is neither a local maximum nor a local minimum (it could be an inflection point, which we'll discuss later).

Example (Continuing from above):

• At x = -1, f'(x) changes from positive to negative. Therefore, f(x) has a local maximum at x = -1. The y-value of this local maximum is f(-1) = (-1)³ - 3(-1)² - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10. So, the local maximum is at the point (-1, 10).
• At x = 3, f'(x) changes from negative to positive. Therefore, f(x) has a local minimum at x = 3. The y-value of this local minimum is f(3) = (3)³ - 3(3)² - 9(3) + 5 = 27 - 27 - 27 + 5 = -22. So, the local minimum is at the point (3, -22).

Step 6: Consider the Domain of the Function

Always be mindful of the domain of the original function f(x). The intervals of increase and decrease must be within the function's domain. For example, if f(x) = √(x), the domain is x ≥ 0. You wouldn't consider intervals where x < 0.

Step 7: Write Your Conclusion

Clearly state the intervals where the function is increasing and decreasing. Also, identify any local maxima or minima.

Example (Final Conclusion):

For the function f(x) = x³ - 3x² - 9x + 5:

• f(x) is increasing on the interval (-∞, -1).
• f(x) is decreasing on the interval (-1, 3).
• f(x) is increasing on the interval (3, ∞).
• f(x) has a local maximum at the point (-1, 10).
• f(x) has a local minimum at the point (3, -22).

A More Complex Example: Dealing with Discontinuities

Let's tackle a function that presents a bit more of a challenge: f(x) = x / (x² - 1).

Step 1: Find the Derivative

We need to use the quotient rule: If f(x) = u(x) / v(x), then f'(x) = [v(x)u'(x) - u(x)v'(x)] / [v(x)]².

• u(x) = x => u'(x) = 1
• v(x) = x² - 1 => v'(x) = 2x

Therefore, f'(x) = [(x² - 1)(1) - (x)(2x)] / (x² - 1)² = (x² - 1 - 2x²) / (x² - 1)² = (-x² - 1) / (x² - 1)² = -(x² + 1) / (x² - 1)²

Step 2: Find Critical Points

• Set the derivative equal to zero: -(x² + 1) / (x² - 1)² = 0. This implies that -(x² + 1) = 0, or x² = -1. This equation has no real solutions. Therefore, there are no critical points where the derivative is zero.
• Identify points where the derivative is undefined: The derivative is undefined when the denominator is zero: (x² - 1)² = 0. This means x² - 1 = 0, so x² = 1, and x = ±1. These are points of discontinuity in the original function, not critical points in the traditional sense, but we must include them on our sign chart because they divide the number line and might indicate changes in increasing/decreasing behavior. The function f(x) is also undefined at these points.

Step 3: Create a Sign Chart

Our points of discontinuity are x = -1 and x = 1. This divides the number line into three intervals:

• Interval 1: (-∞, -1)
• Interval 2: (-1, 1)
• Interval 3: (1, ∞)

Let's choose test values:

• Interval 1: Let's pick x = -2. f'(-2) = -((-2)² + 1) / ((-2)² - 1)² = -(4 + 1) / (4 - 1)² = -5 / 9. Since f'(-2) < 0, the derivative is negative in this interval.
• Interval 2: Let's pick x = 0. f'(0) = -(0² + 1) / (0² - 1)² = -(0 + 1) / (0 - 1)² = -1 / 1 = -1. Since f'(0) < 0, the derivative is negative in this interval.
• Interval 3: Let's pick x = 2. f'(2) = -(2² + 1) / (2² - 1)² = -(4 + 1) / (4 - 1)² = -5 / 9. Since f'(2) < 0, the derivative is negative in this interval.

Our sign chart looks like this:

        (-∞, -1)    (-1, 1)    (1, ∞)
---------------------------------------
f'(x):      -          -          -

Step 4: Determine Intervals of Increase and Decrease

Based on our sign chart:

• f(x) is decreasing on the interval (-∞, -1) because f'(x) < 0.
• f(x) is decreasing on the interval (-1, 1) because f'(x) < 0.
• f(x) is decreasing on the interval (1, ∞) because f'(x) < 0.

Step 5: Identify Local Maxima and Minima

Since the derivative never changes sign at x = -1 or x = 1 (and these aren't even points in the domain of the original function), there are no local maxima or minima. The function simply decreases on either side of these vertical asymptotes.

Step 6: Consider the Domain of the Function

The domain of f(x) = x / (x² - 1) is all real numbers except x = -1 and x = 1. Our intervals reflect this restriction.

Step 7: Write Your Conclusion

For the function f(x) = x / (x² - 1):

• f(x) is decreasing on the interval (-∞, -1).
• f(x) is decreasing on the interval (-1, 1).
• f(x) is decreasing on the interval (1, ∞).
• f(x) has no local maxima or minima. It has vertical asymptotes at x = -1 and x = 1.

Important Considerations and Common Mistakes

• Don't confuse f(x) and f'(x): Remember that f(x) gives you the y-value of the function, while f'(x) gives you the slope of the function.

• Pay attention to endpoints: If the interval is closed (e.g., [a, b]), you need to consider the behavior of the function at the endpoints a and b. The function might be increasing or decreasing up to (or from) the endpoint.

• Undefined derivatives vs. critical points: A point where the derivative is undefined is not necessarily a critical point. Critical points are where f'(x) = 0 or f'(x) is undefined and the point is in the domain of the original function. As we saw in the second example, points where the derivative is undefined often indicate vertical asymptotes or other discontinuities. These points must be included in your sign chart, but you shouldn't call them "critical points."

• Second Derivative Test: While the first derivative test uses the sign of f'(x) to determine increasing/decreasing behavior and local extrema, the second derivative test uses the sign of f''(x) (the second derivative) to determine concavity (whether the function is curving upwards or downwards). The second derivative test can sometimes be used to confirm whether a critical point is a local max or min, but it's not always applicable (e.g., if f''(x) = 0 at the critical point). The first derivative test is generally more reliable.

• Inflection Points: An inflection point is a point on the curve where the concavity changes (from concave up to concave down, or vice versa). Inflection points occur where the second derivative f''(x) = 0 or is undefined. Finding inflection points requires a similar process to finding intervals of increase and decrease, but using the second derivative instead of the first.

Applications of Intervals of Increase and Decrease

Understanding intervals of increase and decrease isn't just an abstract mathematical exercise. It has numerous practical applications:

• Optimization Problems: Many real-world problems involve finding the maximum or minimum value of a function (e.g., maximizing profit, minimizing cost). By identifying intervals of increase and decrease, you can pinpoint potential locations of maxima and minima.

• Curve Sketching: Knowing where a function is increasing, decreasing, and has local extrema is invaluable for creating an accurate sketch of its graph.

• Modeling Real-World Phenomena: Functions are used to model various phenomena in physics, engineering, economics, and other fields. Analyzing intervals of increase and decrease can provide insights into the behavior of these models. For example, you could model the population growth of a species or the spread of a disease.

Conclusion

Mastering the art of finding intervals of increase and decrease is a vital step in your calculus journey. By understanding the relationship between the derivative and the function's behavior, you can unlock a wealth of information about its shape, its turning points, and its overall characteristics. Remember to practice these steps diligently, and you'll be well-equipped to tackle even the most challenging calculus problems. Remember to always carefully consider the domain of the function and to distinguish between critical points and points where the derivative is undefined. With practice, you'll develop a strong intuition for how functions behave and how their derivatives reveal their secrets.