On What Interval Is F Increasing

Let's delve into the concept of determining the intervals where a function, denoted as f, is increasing. This is a fundamental topic in calculus and analysis, crucial for understanding the behavior and properties of functions. Finding these intervals involves analyzing the function's derivative and its sign.

Understanding Increasing Functions

A function f is said to be increasing on an interval I if, for any two points x₁ and x₂ in I such that x₁ < x₂, we have f(x₁) < f(x₂). In simpler terms, as the input x increases, the output f(x) also increases. Geometrically, this means that the graph of the function is rising as you move from left to right.

The Role of the Derivative

The derivative of a function, denoted as f'(x), provides vital information about the function's rate of change. Here's the key connection:

• If f'(x) > 0 on an interval I, then f is increasing on I.
• If f'(x) < 0 on an interval I, then f is decreasing on I.
• If f'(x) = 0 on an interval I, then f is constant on I.

This relationship forms the basis for determining where a function is increasing or decreasing. The derivative acts as a slope indicator; a positive derivative indicates a positive slope (increasing function), a negative derivative indicates a negative slope (decreasing function), and a zero derivative indicates a horizontal line (constant function).

Steps to Determine Intervals of Increase

To find the intervals where a function f(x) is increasing, follow these steps:

1. Find the derivative: Calculate the derivative of the function, f'(x). This is the first and most crucial step. 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.

2. Find critical points: Determine the critical points of the function. Critical points are the values of x where f'(x) = 0 or f'(x) is undefined. These points are crucial because they often mark the boundaries between intervals where the function is increasing or decreasing. Set the derivative equal to zero and solve for x. Also, identify any values of x where the derivative is undefined (e.g., due to division by zero or a square root of a negative number).

3. Create a number line: Draw a number line and mark all the critical points you found in the previous step. These critical points divide the number line into several intervals.

4. Test intervals: Choose a test value x within each interval on the number line. Evaluate the derivative f'(x) at each test value. The sign of f'(x) will indicate whether the function is increasing or decreasing on that interval.

   • If f'(x) > 0 for a test value in the interval, then f is increasing on that interval.
   • If f'(x) < 0 for a test value in the interval, then f is decreasing on that interval.
   • If f'(x) = 0 for a test value in the interval, then f is neither increasing nor decreasing (constant) at that point.

5. State the intervals: Based on the results of the test intervals, state the intervals where the function f(x) is increasing. Express these intervals using interval notation.

Example 1: A Simple Polynomial

Let's find the intervals where the function f(x) = x³ - 3x² + 2 is increasing.

1. Find the derivative:

   • f'(x) = 3x² - 6x

2. Find critical points:

   • Set f'(x) = 0: 3x² - 6x = 0
   • Factor out 3x: 3x(x - 2) = 0
   • Solve for x: x = 0 or x = 2
   • The derivative is defined for all real numbers, so there are no points where f'(x) is undefined.

3. Create a number line:

   • Draw a number line and mark the critical points x = 0 and x = 2. This divides the number line into three intervals: (-∞, 0), (0, 2), and (2, ∞).

4. Test intervals:

   • Interval (-∞, 0): Choose x = -1 as a test value.

     • f'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9 > 0
     • Therefore, f is increasing on the interval (-∞, 0).

   • Interval (0, 2): Choose x = 1 as a test value.

     • f'(1) = 3(1)² - 6(1) = 3 - 6 = -3 < 0
     • Therefore, f is decreasing on the interval (0, 2).

   • Interval (2, ∞): Choose x = 3 as a test value.

     • f'(3) = 3(3)² - 6(3) = 27 - 18 = 9 > 0
     • Therefore, f is increasing on the interval (2, ∞).

5. State the intervals:

   • The function f(x) = x³ - 3x² + 2 is increasing on the intervals (-∞, 0) and (2, ∞).

Example 2: A Rational Function

Let's analyze the function f(x) = (x + 1) / (x - 2) and determine its intervals of increase.

1. Find the derivative: We'll use the quotient rule: f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]²

   • Let u(x) = x + 1 and v(x) = x - 2
   • u'(x) = 1 and v'(x) = 1
   • f'(x) = [(1)(x - 2) - (x + 1)(1)] / (x - 2)² = (x - 2 - x - 1) / (x - 2)² = -3 / (x - 2)²

2. Find critical points:

   • Set f'(x) = 0: -3 / (x - 2)² = 0. This equation has no solution, as -3 can never equal zero.
   • However, f'(x) is undefined when the denominator is zero: (x - 2)² = 0 => x = 2. Therefore, x = 2 is a critical point (where the derivative is undefined).

3. Create a number line:

   • Draw a number line and mark the critical point x = 2. This divides the number line into two intervals: (-∞, 2) and (2, ∞).

4. Test intervals:

   • Interval (-∞, 2): Choose x = 0 as a test value.

     • f'(0) = -3 / (0 - 2)² = -3 / 4 < 0
     • Therefore, f is decreasing on the interval (-∞, 2).

   • Interval (2, ∞): Choose x = 3 as a test value.

     • f'(3) = -3 / (3 - 2)² = -3 / 1 = -3 < 0
     • Therefore, f is decreasing on the interval (2, ∞).

5. State the intervals:

   • The function f(x) = (x + 1) / (x - 2) is never increasing. It is decreasing on the intervals (-∞, 2) and (2, ∞). Note that x = 2 is not included in either interval because the function is undefined at that point.

Example 3: A Trigonometric Function

Let's consider f(x) = sin(x) on the interval [0, 2π] and find where it's increasing.

1. Find the derivative:

   • f'(x) = cos(x)

2. Find critical points:

   • Set f'(x) = 0: cos(x) = 0
   • On the interval [0, 2π], the solutions are x = π/2 and x = 3π/2.

3. Create a number line:

   • Draw a number line representing the interval [0, 2π] and mark the critical points x = π/2 and x = 3π/2. This divides the interval into three sub-intervals: [0, π/2), (π/2, 3π/2), and (3π/2, 2π].

4. Test intervals:

   • Interval (0, π/2): Choose x = π/4 as a test value.

     • f'(π/4) = cos(π/4) = √2 / 2 > 0
     • Therefore, f is increasing on the interval (0, π/2).

   • Interval (π/2, 3π/2): Choose x = π as a test value.

     • f'(π) = cos(π) = -1 < 0
     • Therefore, f is decreasing on the interval (π/2, 3π/2).

   • Interval (3π/2, 2π): Choose x = 7π/4 as a test value.

     • f'(7π/4) = cos(7π/4) = √2 / 2 > 0
     • Therefore, f is increasing on the interval (3π/2, 2π).

5. State the intervals:

   • The function f(x) = sin(x) is increasing on the intervals (0, π/2) and (3π/2, 2π) within the interval [0, 2π].

Considerations and Caveats

• Endpoints: When considering a closed interval, like [a, b], we need to be careful about the endpoints. The function is increasing at the endpoint a if f'(a) > 0 (from the right) and at the endpoint b if f'(b) > 0 (from the left). This is because we are only concerned with the behavior of the function within the interval.

• Discontinuities: If the function has any discontinuities within the interval, we need to consider them as potential breaking points. The intervals of increase or decrease should not include points of discontinuity.

• Second Derivative Test: While the first derivative test tells us about increasing and decreasing intervals, the second derivative test (f''(x)) can tell us about concavity (whether the function is curving upwards or downwards). If f'(x) = 0 and f''(x) > 0, then f(x) has a local minimum at that point. If f'(x) = 0 and f''(x) < 0, then f(x) has a local maximum at that point.

• Domain: Always consider the domain of the original function f(x). The intervals of increase/decrease must lie within the function's domain.

Importance and Applications

Determining the intervals where a function is increasing is a crucial skill in various fields:

• Optimization: Identifying intervals of increase and decrease helps find local and global maxima and minima of functions, which is fundamental in optimization problems (e.g., maximizing profit, minimizing cost).

• Curve Sketching: Knowing where a function is increasing or decreasing is essential for accurately sketching its graph.

• Physics: In physics, this concept is used to analyze the motion of objects (e.g., when is the velocity of an object increasing?).

• Economics: Economists use these techniques to study supply and demand curves, cost functions, and other economic models.

• Engineering: Engineers apply these concepts to analyze system performance, optimize designs, and solve various engineering problems.

Conclusion

Finding the intervals on which a function is increasing is a fundamental calculus skill with far-reaching applications. By understanding the relationship between the derivative and the function's behavior, and by following the steps outlined above, you can confidently determine these intervals for a wide range of functions. Remember to pay attention to critical points, discontinuities, and the domain of the function to ensure accurate results. Mastering this concept will significantly enhance your ability to analyze and understand the behavior of mathematical functions in various contexts.