Problem 13 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the critical numbers and the open intervals on which the function is increasing or decreasing. Then use a graphing utility to graph the function. $$ y=x^{2}-6 x $$

Step-by-Step Solution

Answer

The critical number for the equation y = x^2 - 6x is 3. The function is decreasing in the interval (-Infinity, 3) and increasing in the interval (3, Infinity).

Step 1: Find the derivative

We start by computing the derivative of the function \( y = x^{2} - 6x \). The derivative, \( y' \), using power rule is, \( y' = 2x - 6 \).

Step 2: Find the critical numbers

Critical numbers are values of x that make the derivative of our function equal to zero or undefined. In our case, the derivative \( y' = 2x - 6 \) is never undefined, so we set it equal to zero and solve for x: \( 2x - 6 = 0 \) => \( 2x = 6 \) => \( x = 3 \). So, the only critical number is 3.

Step 3: Determine the intervals of increase or decrease

We now determine where our function is increasing and where it is decreasing. We do this by testing an input number from each of the intervals determined by our critical number (x<3 and x>3). Let's take 1 for x<3, and 4 for x>3. For x = 1, y' = 2(1) - 6 = -4. For x = 4, y' = 2(4) - 6 = 2. As the derivative is negative when x < 3 and positive when x > 3, it implies that our function is decreasing for x < 3 and increasing for x > 3.

Step 4: Graph the function

Use a graphing utility to plot \( y = x^{2} - 6x \). The point on the graph where the function changes from decreasing to increasing corresponds to x = 3. This verifies our calculation that x = 3 is the critical number.

Key Concepts

Understanding Critical Numbers

Critical numbers of a function are specific values of x where something important happens. These are x values where:

• The derivative of the function is zero, or
• The derivative does not exist (undefined).

It's crucial to identify critical numbers because these points often signal a change in the behavior of a function.
For the function provided, the derivative is always defined, thus the critical numbers come from values where the derivative equals zero.
For our example, the critical number is found by solving the equation \(2x - 6 = 0\), resulting in \(x = 3\). This means something notable is happening in the function at this x-value!

The Role of Derivatives in Calculus

The concept of a derivative is foundational in calculus. A derivative tells us how a function is changing at any particular point.
It's like a speedometer that shows the rate of change of the function's output as the input x changes. The general steps to find a derivative include:

• Identify the form of the original function.
• Apply the appropriate derivative rules, such as the power rule, product rule, or chain rule.

In our specific function, \(y = x^2 - 6x\), we used the power rule to find that the derivative is \(y' = 2x - 6\). This reveals the rate of change of the function and all important characteristics like where it may increase or decrease.
Finding the derivative gives us deeper insights into the behavior and characteristics of the function itself.

Exploring Increasing and Decreasing Functions

Increasing and decreasing intervals describe where a function goes up or down as you move along the x-axis.
An interval where a function is increasing means every output (y-value) is getting larger as the input (x-value) increases. Conversely, in a decreasing interval, the y-values are getting smaller.
Once we calculate the critical number, \(x = 3\) in our case, we check the derivative on either side of this point:

• For \(x < 3\), using \(x = 1\), we found \(y' = -4\), indicating that the function is decreasing there.
• For \(x > 3\), using \(x = 4\), \(y' = 2\) shows the function is increasing in this range.

This analysis helps us to conclude that the function decreases from \(-\infty\) to 3, and increases from 3 to \(\infty\). Understanding these behaviors is crucial for sketching the graph and predicting how the function behaves globally.