/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 In Exercises 29-32, graph the fu... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 29

In Exercises 29-32, graph the functions \(f\), \(g\), and \(f\) + \(g\) on the same set of coordinate axes. \(f(x) = \frac{1}{2}x\), \(g(x) = x - 1\)

Short Answer

Expert verified
The graph of the function \(f(x) = \frac{1}{2}x\) is a line with a slope of 0.5 through the origin. The graph of the function \(g(x) = x - 1\) is a line with a slope of 1 and y-intercept -1. The graph of function \(f(x) + g(x) = \frac{3}{2}x - 1\) is a line with a slope of 1.5 and y-intercept -1.

Step by step solution

01

Graph function \(f(x)\)

The graph of function \(f(x) = \frac{1}{2}x\) is a straight line with a slope of 0.5 passing through the origin. Choose several x-values, amounting to varied points, then draw a line through these points to represent the function graph.
02

Graph function \(g(x)\)

The graph of function \(g(x) = x - 1\) is a straight line with a slope of 1 and y-intercept of -1. Like the previous step, choose several x-values, sum up the corresponding y-values, and then draw a line through these points to represent the function graph.
03

Graph \(f(x) + g(x)\)

The next function to graph is \(f(x) + g(x)\), which is the sum of \(f(x)\) and \(g(x)\). To get this, add the y-values of functions \(f(x)\) and \(g(x)\) for every x. This translates into the function \( f(x) + g(x) = \frac{1}{2}x + (x - 1) = \frac{3}{2}x - 1\). The graph of this function is a straight line with a slope of 1.5 and y-intercept of -1. Perform the same procedures as in the previous steps particularly choosing values for x, summing up the corresponding y-values, and then drawing a line through these points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
Understanding the slope-intercept form is essential when graphing linear functions. It's the formula represented as \( y = mx + b \<\/strong>\) where \\) stands for the slope of the line and \\( b \<\/strong>\) represents the y-intercept. The y-intercept is simply where the line crosses the y-axis.

In the exercise, the function \( f(x) \<\/strong>\) can be written as \( y = \frac{1}{2}x \<\/strong>\) in slope-intercept form. It has a slope (m) of \\( \frac{1}{2} \<\/strong>\) and a y-intercept (b) of 0 since it passes through the origin. On the other hand, for the function \( g(x) \<\/strong>\) which is expressed as \( y = x - 1 \<\/strong>\) , the slope is 1 and the y-intercept is -1. Understanding the slope and y-intercept helps you to quickly graph these linear functions.

For better understanding, try finding two points that satisfy the equation. For example, for the function \\( g(x) \<\/strong>\) , when \( x = 1 \<\/strong>\) , the \( y = 0 \<\/strong>\) and when \( x = 2 \<\/strong>\) , \( y = 1 \<\/strong>\) . Draw a straight line through these two points, and you have the graph of \\( g(x) \<\/strong>\) .
Sum of Functions
When dealing with the sum of functions, you simply add the corresponding function values (\( y \<\/strong>\) -values) for each \\( x \<\/strong>\) -value. The exercise presents two functions, \\( f(x) \<\/strong>\) and \\( g(x) \<\/strong>\) , and requires you to graph the sum \( f(x) + g(x) \<\/strong>\) . Let's break it down:

Firstly, calculate the sum algebraically by adding \\( f(x) \<\/strong>\) and \\( g(x) \<\/strong>\) . In this case, \\( \frac{1}{2}x + (x - 1) \<\/strong>\) simplifies to \\( \frac{3}{2}x - 1 \<\/strong>\) . This new equation is now your function to graph, which already is in the slope-intercept form with a slope of \\( \frac{3}{2} \<\/strong>\) and y-intercept of -1.

Graphically, for each \\( x \<\/strong>\) -value, you would calculate both \\( f(x) \<\/strong>\) and \\( g(x) \<\/strong>\) , then add them. If you take \\( x = 2 \<\/strong>\) , for instance, \\( f(2) = \frac{1}{2} \times 2 = 1 \<\/strong>\) and \\( g(2) = 2 - 1 = 1 \<\/strong>\) , hence \\( f(2) + g(2) = 1 + 1 = 2 \<\/strong>\) . Do this for a few \\( x \<\/strong>\) -values, plot the resulting points, draw a line through them, and you will have visually represented the sum of the two functions.
Graphing Techniques
Effective graphing techniques can make all the difference in how well students grasp the visual representation of functions. The step-by-step solutions already gave us a sneak peek into some of these techniques.

When graphing a linear function like \( f(x) \<\/strong>\) or \( g(x) \<\/strong>\) , you can start by plotting the y-intercept. From there, use the slope to determine the direction in which the line will continue. Remember, the slope ratio \\( m = \frac{\text{rise}}{\text{run}} \<\/strong>\) tells you how many units to go up or down for every unit you go right or left.

For the equation \( \frac{1}{2}x \<\/strong>\) , this means you go up 1 unit for every 2 units you move to the right since the slope is \\( \frac{1}{2} \<\/strong>\) . And for the function \\( x - 1 \<\/strong>\) , the slope is 1, which means you move up 1 unit for each unit you move to the right.

Another helpful technique is selecting points that are easy to calculate and plot. For a more accurate graph, it's advisable to choose at least three points to ensure your graph is straight and follows the correct slope. Finally, connect the plotted points with a ruler to ensure a straight line, completing the visual representation of the linear function.

Learning to confidently implement these graphing techniques can enable you to graph any linear function quickly and effectively.

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Most popular questions from this chapter

In Exercises 109-112, determine if the situation could be represented by a one-to-one function. If so, write a statement that describes the inverse function. The height \(h\) in inches of a human born in the year 2000 in terms of his or her age \(n\) in years.

In Exercises 63-76, determine whether the function has an inverse function. If it does, find the inverse function. \(g(x) = (x+3)^2\), \(x \geq -3\)

HOURLY WAGE Your wage is \(\$10.00\) per hour plus \(\$0.75\) for each unit produced per hour. So, your hourly wage \(y\) in terms of the number of units produced \(x\) is \(y = 10 + 0.75x\). (a) Find the inverse function. What does each variable represent in the inverse function? (b) Determine the number of units produced when your hourly wage is \(\$24.25\).

SPORTS The lengths (in feet) of the winning men's discus throws in the Olympics from 1920 through 2008 are listed below. (Source: International Olympic Committee) 1920 146.6 1924 151.3 1928 155.3 1932 162.3 1936 165.6 1948 173.2 1952 180.5 1956 184.9 1960 194.2 1964 200.1 1968 212.5 1972 211.3 1976 221.5 1980 218.7 1984 218.5 1988 225.8 1992 213.7 1996 227.7 2000 227.3 2004 229.3 2008 225.8 (a) Sketch a scatter plot of the data. Let \(y\) represent the length of the winning discus throw (in feet) and let \(t=20\) represent 1920. (b) Use a straightedge to sketch the best-fitting line through the points and find an equation of the line. (c) Use the \(regression\) feature of a graphing utility to find the least squares regression line that fits the data. (d) Compare the linear model you found in part (b) with the linear model given by the graphing utility in part (c). (e) Use the models from parts (b) and (c) to estimate the winning men's discus throw in the year 2012.

THINK ABOUT IT In Exercises 77-86, restrict the domain of the function \(f\) so that the function is one-to-one and has an inverse function. Then find the inverse function \(f^{-1}\). State the domains and ranges of \(f\) and \(f^{-1}\). Explain your results. (There are many correct answers.) \(f(x) = -|x-1| - 2\)
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