Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use a computer or a graphing calculator Let \(f(x)=1 /\left(x^{2}+1\right) .\) Using the same axes, draw the graphs of \(y=f(x), y=f(2 x)\), and \(y=f(x-2)+0.6\), all on the domain \([-4,4]\).

Step-by-Step Solution

Answer

Three functions: \(y=f(x)\) is symmetric, \(y=f(2x)\) is horizontally compressed, \(y=f(x-2)+0.6\) is shifted right and up.

Step 1: Understand the base function

The base function is given by \(f(x) = \frac{1}{x^2 + 1}\). This function is symmetric with respect to the y-axis and has a horizontal asymptote at \(y = 0\). The function never touches the x-axis and reaches its maximum value of 1 at \(x=0\).

Step 2: Transform the base function to obtain \(y=f(2x)\)

The transformation \(f(2x)\) is a horizontal compression of \(f(x)\) by a factor of 2. This means that the function \(y=f(2x)\) will be skinnier compared to \(y=f(x)\). Points on the graph will be closer to the y-axis.

Step 3: Perform another transformation to sketch \(y=f(x-2)+0.6\)

The transformation \(f(x-2)\) shifts the base function \(f(x)\) 2 units to the right, and the additional \(+0.6\) shifts the entire function up by 0.6 units. Hence, \(y=f(x-2)+0.6\) will be the graph of \(f(x)\) moved right by 2 units and then up by 0.6 units.

Step 4: Use technology to graph the functions

To graph these functions, use graphing software or a calculator. Set up the domain from \(-4\) to \(4\), and graph \(y = f(x)\), \(y = f(2x)\), and \(y = f(x-2) + 0.6\) on the same set of axes. Observe the differences in shape and position of the graphs.

Key Concepts

Horizontal Compression

When we talk about horizontal compression in graph transformations, we're essentially "squeezing" the graph along the x-axis. Imagine the graph being pushed inward from both sides. For example, with the function given in the exercise, the transformation from \( y = f(x) \) to \( y = f(2x) \) is a horizontal compression. This means every x-value of the function is halved.
It's like the x-values are getting pulled closer to the y-axis.
Instead of using each x-value, now we use twice as many to get the same position in the graph.

• This makes the graph appear narrower or skinnier.
• The y-values reach their max or min faster compared to the original function.

Understanding horizontal compression helps you predict how a graph will change just by looking at the function formula.

Vertical Shift

A vertical shift involves moving the entire graph of a function up or down, without altering its shape. In the transformation \(y = f(x-2) + 0.6\) from the exercise, the part '+0.6' is responsible for shifting the graph upward.
Every point on the graph moves 0.6 units higher on the y-axis.

• This doesn’t change where the function hits along the x-axis.
• It's an easy way to adjust the height of a graph.

The key to spotting a vertical shift is looking for something added or subtracted from the function, outside of the parenthesis. By changing the vertical position, you can see relative differences in how different graphs relate to each other.

Domain and Range

When analyzing functions like \(f(x)\), understanding domain and range is crucial. The **domain** is all the x-values a function can accept, while the **range** is all the possible y-values a function can produce.
For the base function \(f(x) = \frac{1}{x^2 + 1}\), its domain is all real numbers, \([-\infty, \infty]\), since you can plug any real number into \(x^2 + 1\) without causing any mathematical issues.
The range is \((0, 1]\) because the fraction \(\frac{1}{x^2 + 1}\) never outputs a value less than 0 and reaches a maximum of 1 at \(x = 0\).

• The horizontal compression in \(f(2x)\) doesn't alter the domain or range since the function itself doesn't change vertically.
• In contrast, the vertical shift in \(f(x-2) + 0.6\) primarily affects the range, moving it up by 0.6, altering how high the function can reach.

Grasping the domain and range enables you to understand the limits within which a graph operates.