Problem 40 - Precalculus with Limits

Question

In Exercises 39-46, determine the intervals over which the function is increasing, decreasing, or constant. \(f(x) = x^2 - 4x\)

Step-by-Step Solution

Answer

The function \(f(x) = x^2 - 4x\) is decreasing on the interval (-∞, 2) and increasing on the interval (2, ∞).

Step 1: Differentiation

First, calculate the first derivative of the function. The derivative of \(f(x) = x^2 - 4x\) is \( f'(x) = 2x - 4 \). This derivative represents the rate of change of the function

Step 2: Find Critical Points

Find the critical points by setting the derivative equal to zero, i.e, \( 2x - 4 = 0 \). Solving this equation, we get the critical point \( x = 2 \). This point divides the real number line into intervals, (-∞, 2) and (2, ∞).

Step 3: Determine Increasing or Decreasing Intervals

Take a test point from each interval and substitute it into the derivative. If the derivative is positive at that point, the function is increasing in that interval. If it's negative, the function is decreasing. For the interval (-∞, 2), let's take x=0 then \( f'(0) = -4 \) which means the function is decreasing in this interval. For the interval (2, ∞), let's take x=3 then \( f'(3) = 2 \) which means the function is increasing in this interval.

Key Concepts

Understanding the Derivative

The derivative of a function is a powerful tool in calculus. It tells us how a function is changing at any given point. In other words, the derivative provides the rate of change of the function's output concerning its input.

Consider the function given: \( f(x) = x^2 - 4x \). To find its derivative, we apply basic differentiation rules. For the term \( x^2 \), the derivative is \( 2x \), and for \( -4x \), the derivative is \( -4 \). Adding these, the first derivative is \( f'(x) = 2x - 4 \).

• The derivative \( 2x - 4 \) helps identify how steep or flat the function is at any point \( x \).
• If \( f'(x) > 0 \), the function is rising at \( x \).
• If \( f'(x) < 0 \), the function is falling at \( x \).

Knowing how to compute and interpret the derivative is the first step in understanding how functions behave.

Identifying Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points are essential because they are candidates for local maxima, minima, or inflection points.

You find critical points by solving \( f'(x) = 0 \). For our function \( f(x) = x^2 - 4x \), we calculated the derivative to be \( 2x - 4 \). Setting \( 2x - 4 = 0 \), we get \( x = 2 \). Now, \( x = 2 \) is a critical point.

• Critical points help breakup the domain into intervals to analyze increasing or decreasing behavior.
• It is not guaranteed that all critical points are where a function switches from increasing to decreasing or vice-versa, but they are often where this occurs.

Understanding critical points and their significance is crucial for analyzing the overall behavior of a function.

Intervals of Increasing and Decreasing Functions

Once you know the critical points, you can determine intervals where the function increases or decreases. These intervals tell us about the overall shape and behavior of the function.

Using the critical point \( x = 2 \), we divide our number line into two intervals: \((-\infty, 2)\) and \((2, \infty)\). We then select test points from these intervals to check the sign of the derivative:

• For \( x = 0 \) in \((-\infty, 2)\), \( f'(0) = -4 \), indicating the function decreases in this interval.
• For \( x = 3 \) in \((2, \infty)\), \( f'(3) = 2 \), indicating the function increases in this interval.

By investigating the sign of \( f'(x) \), we know whether a function goes up or down on specific intervals.

This process helps understand where a function is increasing or decreasing and how that relates to real-world problems or more complex mathematical analyses.