Graphs of Functions
Contents: This page corresponds to § 1.4 (p. 116) of the text.
Suggested Problems from Text
p. 124 #1, 2, 5, 8, 9, 11, 16, 17, 21, 25, 27, 29, 31, 39, 40, 47, 50, 51, 52, 54, 57, 64, 65, 66
Defining the Graph of a Function
Vertical Line Test
Characteristics of Graphs
Approximating Relative Extrema
Even and Odd Functions
The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also define the graph of f to be the graph of the equation y = f(x). So, the graph of a function if a special case of the graph of an equation.
Let f(x) = x2 - 3.
Recall that when we introduced graphs of equations we noted that if we can solve the equation for y, then it is easy to find points that are on the graph. We simply choose a number for x, then compute the corresponding value of y. Graphs of functions are graphs of equations that have been solved for y!
The graph of f(x) in this example is the graph of y = x2 - 3. It is easy to generate points on the graph. Choose a value for the first coordinate, then evaluate f at that number to find the second coordinate. The following table shows several values for x and the function f evaluated at those numbers.
Each column of numbers in the table holds the coordinates of a point on the graph of f.
(a) Plot the five points on the graph of f from the table above, and based on these points, sketch the graph of f.
(b) Verify that your sketch is correct by using the Java Grapher to graph f. Simply enter the formula x^2 - 3 in the f text box and click graph.
Let f be the piecewise-defined function
To express f in a single formula for the Java Grapher or Java Calculator we write
(5 - x^2)*(xLE2) + (x - 1)*(2Lx).
The factor (xLE2) has the value 1 for x <= 2 and 0 for x > 2. Similarly, (2Lx) is 1 for 2 < x and 0 otherwise. If we evaluate the sum above at x = 3, the first product is 0 because (xLE2) is 0 and the second product is (3 - 1)*1=2. In other words, for x > 2, the formula evaluates to x - 1. If x <= 2, then the formula above is equal to 5 - x^2, which is exactly what we want!
The graph of f is shown below.
Graph the piecewise-defined function
We have seen that some equations in x and y do not describe y as a function of x. The algebraic way see if an equation determines y as a function of x is to solve for y. If there is not a unique solution, then y is not a function of x.
Suppose that we are given the graph of the equation. There is an easy way to see if this equation describes y as a function of x.
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A set of points in the plane is the graph of a function if and only if no vertical line intersects the graph in more than one point.
The graph of the equation y2 = x + 5 is shown below.
By the vertical line test, this graph is not the graph of a function, because there are many vertical lines that hit it more than once.
Think of the vertical line test this way. The points on the graph of a function f have the form (x, f(x)), so once you know the first coordinate, the second is determined. Therefore, there cannot be two points on the graph of a function with the same first coordinate.
All the points on a vertical line have the same first coordinate, so if a vertical line hits a graph twice, then there are two points on the graph with the same first coordinate. If that happens, the graph is not the graph of a function. Videos: Animated Gif, MS Avi File, or Real Video File
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Consider the function f(x) = 2 x + 1. We recognize the equation y = 2 x + 1 as the Slope-Intercept form of the equation of a line with slope 2 and y-intercept (0,1).
Think of a point moving on the graph of f. As the point moves toward the right it rises. This is what it means for a function to be increasing. Your text has a more precise definition, but this is the basic idea.
The function f above is increasing everywhere. In general, there are intervals where a function is increasing and intervals where it is decreasing.
The function graphed above is decreasing for x between -3 and 2. It is increasing for x less than -3 and for x greater than 2.
Using interval notation, we say that the function is
decreasing on the interval (-3, 2)
increasing on (-infinity, -3) and (2, infinity)
Graph the function f(x) = x2 - 6x + 7 and find the intervals where it is increasing and where it is decreasing. Answer
Some of the most characteristics of a function are its Relative Extreme Values. Points on the functions graph corresponding to relative extreme values are turning points, or points where the function changes from decreasing to increasing or vice versa. Let f be the function whose graph is drawn below.
f is decreasing on (-infinity, a) and increasing on (a, b), so the point (a, f(a)) is a turning point of the graph. f(a) is called a relative minimum of f. Note that f(a) is not the smallest function value, f(c) is. However, if we consider only the portion of the graph in the circle above a, then f(a) is the smallest second coordinate. Look at the circle on the graph above b. While f(b) is not the largest function value (this function does not have a largest value), if we look only at the portion of the graph in the circle, then the point (b, f(b)) is above all the other points. So, f(b) is a relative maximum of f. f(c) is another relative minimum of f. Indeed, f(c) is the absolute minimum of f, but it is also one of the relative minima.
Here again we are giving definitions that appeal to your geometric intuition. The precise definitions are given in your text.
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Finding the exact location of a function's relative extrema generally requires calculus. However, graphing utilities such as the Java Grapher may be used to approximate these numbers.
Note on terminology:
Suppose a is a number such that f(a) is a relative minimum. In applications, it is often more important to know where the function attains its relative minimum than it is to know what the relative minimum is.
For example, f(x) = x3 - 4x2 + 4x has a relative minimum of 0. It attains this relative minimum at x = 2, so (2,0) is a turning point of the graph of f. We will call the point (2,0) a relative minimum point. In general, a relative extreme point is a point on the graph of f whose second coordinate is a relative extreme value of f.
Approximate the relative extremes points of f(x) = x3 - x2 - 6x.
When you display the graph of f in the default viewing rectangle you see that f has one relative maximum point near (-1,4) and one relative minimum point near (2,-8). The approximations (-1,4) and (2,-8) are not very close to the real relative extreme points, so we will use the zoom and trace features to improve the approximations.
When you click the Trace button, a point on the graph of f is indicated with a small circle. The coordinates of that point are reported in the two text boxes near the Trace button. As you click the box with the right arrow >, the trace point moves to the right, staying on the graph of f. If you select a larger Step Size from the pull down menu, then the trace point moves farther with each click. Also notice that once you have clicked the > button, then you can use the enter key to move further right. It is possible to move faster with the enter key than with the mouse.
Using the default view, the lowest point found while tracing near the minimum point is (1.8, -8.208). Note that this is not the exact location of the minimum point. We need to look at the trace points on either side of this point to get an idea of how close we are. Find this trace point, make sure that the Step Size is set to 1, and then find the points on either side of this point. The table below lists the coordinates of these points.
1.7333333 -8.196741 1.8 -8.208 1.8666667 -8.180148
Note that different Java implementations will compute and report different accuracies, so the values you find may may be slightly different from those above.
The points in the table show that the real minimum point has an x coordinate somewhere between 1.7333333 and 1.8666667. Note that we do not yet have enough information to report the x value with even one decimal place accuracy, because if the second decimal place were a 4, then the value would round to 1.7. If the second decimal place were a 7, then the value would round to 1.8. So we need to improve our estimate by zooming in on the minimum point. There are several ways to do this with the Grapher: Zoom In, Zoom Box, or set the coordinates of the viewing rectangle. In this example we will set the view coordinates.
Type in Xmin = 1.74, Xmax = 1.86, Ymin = -8.21, Ymax = -8.19, and set the viewing rectangle to these coordinates by clicking the Reset button. Changing the viewing rectangle removes the trace point. To get it back, click the Trace button again. Now you need to move left to get to the minimum point. There are two different x values that correspond to the lowest y value.
1.786 -8.20882 1.7864 -8.20882
We still do not know the exact location of the minimum point, but we know that its x coordinate is between 1.786 and 1.7864. That means the x coordinate will be 1.786 after rounding to three decimal places, . Since we are only reporting three decimal places for the x coordinate, we will also round the y coordinate to three decimal places, so our approximation is (1.786, -8.209).
Find the relative maximum point to two decimal place accuracy. Answer
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A function f is even if its graph is symmetric with respect to the y-axis. This criterion can be stated algebraically as follows: f is even if f(-x) = f(x) for all x in the domain of f. For example, if you evaluate f at 3 and at -3, then you will get the same value if f is even.
This condition is very easy to check with the Java Grapher.
Open the Grapher and type (x - 2)^2 into the f text box and click the Graph button. This function is not even, so when we graph its reflection about the y-axis, we will get a new graph.
In the g text box type f(-x), and click the Graph button The graph of g is the reflection about the y-axis of the graph of f. Since we see two distinct graphs, we know that f is not even.
Now replace the text in the f box with x^2 - 3; clear the text from the g box and graph the function. This graph is symmetric with respect to the y-axis, so when you enter f(-x) in the g box and graph again, you do not see anything new. This is because the graph of g is the same as the graph of f.
A function f is odd if its graph is symmetric with respect to the origin. This criterion can be stated algebraically as follows: f is odd if f(-x) = -f(x) for all x in the domain of f. For example, if you evaluate f at 3, you get the negative of f(-3) when f is odd.
If you enter any function in the f box of the Grapher and enter -f(-x) in the g box, then the graph of g is the reflection through the origin of the graph of f. So, if f is not odd, then you see two distinct graphs. If f is odd, you see only one graph.
Graph f(x) = (x - 2)^2 and g(x) = -f(-x). Because you see two distinct graphs, f is not odd.
Now enter f(x) = x*(x^2 - 1) and "turn off" the graph of g by unchecking the box to the right of the text box and click Graph again. With this function f, the graph of g is the same as the graph of f. So, when you turn on the graph of g and click Graph again, you see nothing new.
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