# 1.7 Graphs of Functions

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1 64 Relations and Functions 1.7 Graphs of Functions In Section 1.4 we defined a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Then in Section 1.5 we described a function as a process and defined the notation necessary to work with functions algebraically. So now it s time to look at functions graphically again, only this time we ll do so with the notation defined in Section 1.5. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions The graph of a function f is the set of points which satisfy the equation y = f(x). That is, the point (x, y) is on the graph of f if and only if y = f(x). Example Graph f(x) = x 2 x 6. Solution. To graph f, we graph the equation y = f(x). To this end, we use the techniques outlined in Section 1.3. Specifically, we check for intercepts, test for symmetry, and plot additional points as needed. To find the x-intercepts, we set y = 0. Since y = f(x), this means f(x) = 0. f(x) = x 2 x 6 0 = x 2 x 6 0 = (x 3)(x + 2) factor x 3 = 0 or x + 2 = 0 x = 2, 3 So we get ( 2, 0) and (3, 0) as x-intercepts. To find the y-intercept, we set x = 0. Using function notation, this is the same as finding f(0) and f(0) = = 6. Thus the y-intercept is (0, 6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.3, plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. x f(x) (x, f(x)) 3 6 ( 3, 6) 2 0 ( 2, 0) 1 4 ( 1, 4) 0 6 (0, 6) 1 6 (1, 6) 2 4 (2, 4) 3 0 (3, 0) 4 6 (4, 6) y x

2 1.7 Graphs of Functions 65 Graphing piecewise-defined functions is a bit more of a challenge. Example Graph: f(x) = { 4 x 2 if x < 1 x 3, if x 1 Solution. We proceed as before: finding intercepts, testing for symmetry and then plotting additional points as needed. To find the x-intercepts, as before, we set f(x) = 0. The twist is that we have two formulas for f(x). For x < 1, we use the formula f(x) = 4 x 2. Setting f(x) = 0 gives 0 = 4 x 2, so that x = ±2. However, of these two answers, only x = 2 fits in the domain x < 1 for this piece. This means the only x-intercept for the x < 1 region of the x-axis is ( 2, 0). For x 1, f(x) = x 3. Setting f(x) = 0 gives 0 = x 3, or x = 3. Since x = 3 satisfies the inequality x 1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x = 0 falls in the domain x < 1. Thus f(0) = = 4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 x 2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x 3 possesses none of the symmetries discussed in the Section 1.3. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x = 1, since this is where the equations change. When x = 1, we use the formula f(x) = x 3, so the point on the graph (1, f(1)) is (1, 2). However, for all values less than 1, we use the formula f(x) = 4 x 2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9, x = 0.99, x = 0.999, and so on, we find the corresponding y values using the formula f(x) = 4 x 2. Making a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f(x) values inch closer and closer 1 to = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get 4 3 y x f(x) (x, f(x)) (0.9, 3.19) (0.99, 3.02) (0.999, 3.002) x 1 We ve just stepped into Calculus here!

4 1.7 Graphs of Functions f(x) = f( x) = f( x) = f( x) = f(x) ( x) 2 5 Hence, f is even. The graphing calculator furnishes the following: This suggests 4 the graph of f is symmetric about the y-axis, as expected. 2. g(x) = 5x g( x) = 5( x) 2 ( x) 2 g( x) = 5x It doesn t appear that g( x) is equivalent to g(x). To prove this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g( 1) = 5. This proves that g is not even, but it doesn t rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g( x) with g(x) Hence, g is odd. Graphically, 4 Suggests is about the extent of what it can do. g(x) = 5x = 5x g(x) = g( x)

5 68 Relations and Functions The calculator indicates the graph of g is symmetric about the origin, as expected. 3. h(x) = 5x 2 x 3 h( x) = 5( x) 2 ( x) 3 h( x) = 5x 2 + x 3 Once again, h( x) doesn t appear to be equivalent to h(x). We check with an x value, for example, h(1) = 5 but h( 1) = 5 3. This proves that h is not even and it also shows h is not odd. (Why?) Graphically, The graph of h appears to be neither symmetric about the y-axis nor the origin. 4. i(x) = i( x) = i( x) = 5x 2x x 3 5( x) 2( x) ( x) 3 5x 2x + x 3 The expression i( x) doesn t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i( 1) = 5, it appears that i( x) does, in fact, equal i(x). However, while this suggests i is even, it doesn t prove it. (It does, however,

6 1.7 Graphs of Functions 69 prove i is not odd.) To prove i( x) = i(x), we need to manipulate our expressions for i(x) and i( x) and show they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i( x) the numerator is 5x. To re-write i(x) with a numerator of 5x, we need to multiply its numerator by 1. To keep the value of the fraction the same, we need to multiply the denominator by 1 as well. Thus i(x) = = = 5x 2x x 3 ( 1)5x ( 1) (2x x 3 ) 5x 2x + x 3 Hence, i(x) = i( x), so i is even. The calculator supports our conclusion. 5. j(x) = x 2 x j( x) = ( x) 2 x j( x) = x 2 + x The expression for j( x) doesn t seem to be equivalent to j(x), so we check using x = 1 to get j(1) = and j( 1) = 100. This rules out j being even. However, it doesn t rule out j being odd. Examining j(x) gives j(x) = x 2 x ( j(x) = x 2 x ) j(x) = x 2 + x The expression j(x) doesn t seem to match j( x) either. Testing x = 2 gives j(2) = and j( 2) = , so j is not odd, either. The calculator gives:

7 70 Relations and Functions The calculator suggests that the graph of j is symmetric about the y-axis which would imply that j is even. However, we have proven that is not the case. There are two lessons to be learned from the last example. The first is that sampling function values at particular x values is not enough to prove that a function is even or odd despite the fact that j( 1) = j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the algebra which proves mathematical truths General Function Behavior The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them all in great detail. The purpose of this section s discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing, and constant. Before defining the concepts algebraically, it is instructive to first look at them graphically. Consider the graph of the function f given on the next page. Reading from left to right, the graph starts at the point ( 4, 3) and ends at the point (6, 5.5). If we imagine walking from left to right on the graph, between ( 4, 3) and ( 2, 4.5), we are walking uphill ; then between ( 2, 4.5) and (3, 8), we are walking downhill ; and between (3, 8) and (4, 6), we are walking uphill once more. From (4, 6) to (5, 6), we level off, and then resume walking uphill from (5, 6) to (6, 5.5). In other words, for the x values between 4 and 2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y = f(x), the y values on the graph are the function values, and we say that the function f is increasing on the interval [ 4, 2]. Analogously, we say that f is decreasing on the interval [ 2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and finally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f is increasing on [ 4, 2] we do not mention the actual y values that f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs. 6 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run 5 Or, in other words, don t rely too heavily on the machine! 6 The notions of how quickly or how slowly a function increases or decreases are explored in Calculus.

8 1.7 Graphs of Functions 71 into serious trouble in our previous example if we tried to do so because x = 2 is contained in an interval on which f was increasing and one on which it is decreasing. (There s more on this issue and many others in the exercises.) ( 2, 4.5) y (6, 5.5) ( 4, 3) 4 5 (4, 6) 6 7 (5, 6) 8 9 (3, 8) x The graph of y = f(x) We re now ready for the more formal algebraic definitions of what it means for a function to be increasing, decreasing or constant. Definition 1.8. Suppose f is a function defined on an interval I. We say f is: increasing on I if and only if f(a) < f(b) for all real numbers a, b in I with a < b. decreasing on I if and only if f(a) > f(b) for all real numbers a, b in I with a < b. constant on I if and only if f(a) = f(b) for all real numbers a, b in I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing, and constant as stated in Definition 1.8 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you find an example of a function for which none of the concepts in Definition 1.8 apply? Now let s turn our attention to a few of the points on the graph. Clearly the point ( 2, 4.5) does not have the largest y value of all of the points on the graph of f indeed that honor goes to

9 72 Relations and Functions (6, 5.5) but ( 2, 4.5) should get some sort of consolation prize for being the top of the hill between x = 4 and x = 3. We say that the function f has a local maximum 7 at the point ( 2, 4.5), because the y-coordinate 4.5 is the largest y-value (hence, function value) on the curve near 8 x = 2. Similarly, we say that the function f has a local minimum 9 at the point (3, 8), since the y-coordinate 8 is the smallest function value near x = 3. Although it is tempting to say that local extrema 10 occur when the function changes from increasing to decreasing or vice versa, it is not a precise enough way to define the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional definitions and thoroughly vet the pathologies they induce in the exercises. We have one last observation to make before we proceed to the algebraic definitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see the largest y value (hence the largest function value) is 5.5 at x = 6. In this case, we say the maximum 11 of f is 5.5; similarly, the minimum 12 of f is 8. We formalize these concepts in the following definitions. Definition 1.9. Suppose f is a function with f(a) = b. We say f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f(a) f(x) for all x in I different than a. The value f(a) = b is called a local maximum value of f in this case. We say f has a local minimum at the point (a, b) if and only if there is an open interval I containing a for which f(a) f(x) for all x in I different than a. The value f(a) = b is called a local minimum value of f in this case. The value b is called the maximum of f if b f(x) for all x in the domain of f. The value b is called the minimum of f if b f(x) for all x in the domain of f. It s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function s graph would have to look like?) We shall see in the exercises examples of functions which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic definitions above as we explore the wonderful diversity of graphs that functions provide to us. Here is the tame example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the exercises. 7 Also called relative maximum. 8 We will make this more precise in a moment. 9 Also called a relative minimum. 10 Maxima is the plural of maximum and mimima is the plural of minimum. Extrema is the plural of extremum which combines maximum and minimum. 11 Sometimes called the absolute or global maximum. 12 Again, absolute or global minimum can be used.

10 1.7 Graphs of Functions 73 Example Given the graph of y = f(x) below, answer all of the following questions. y 4 3 (0, 3) 2 1 ( 2, 0) (2, 0) x ( 4, 3) (4, 3) 1. Find the domain of f. 2. Find the range of f. 3. Determine f(2). 4. List the x-intercepts, if any exist. 5. List the y-intercepts, if any exist. 6. Find the zeros of f. 7. Solve f(x) < Determine the number of solutions to the equation f(x) = List the intervals on which f is increasing. 10. List the intervals on which f is decreasing. 11. List the local maximums, if any exist. 12. List the local minimums, if any exist. 13. Find the maximum, if it exists. 14. Find the minimum, if it exists. 15. Does f appear to be even, odd, or neither? Solution. 1. To find the domain of f, we proceed as in Section 1.4. By projecting the graph to the x-axis, we see the portion of the x-axis which corresponds to a point on the graph is everything from 4 to 4, inclusive. Hence, the domain is [ 4, 4]. 2. To find the range, we project the graph to the y-axis. We see that the y values from 3 to 3, inclusive, constitute the range of f. Hence, our answer is [ 3, 3]. 3. Since the graph of f is the graph of the equation y = f(x), f(2) is the y-coordinate of the point which corresponds to x = 2. Since the point (2, 0) is on the graph, we have f(2) = 0.

11 74 Relations and Functions 4. The x-intercepts are the points on the graph with y-coordinate 0, namely ( 2, 0) and (2, 0). 5. The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). 6. The zeros of f are the x-coordinates of the x-intercepts of the graph of y = f(x) which are x = 2, To solve f(x) < 0, we look for the x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking where the graph is below the x-axis. This happens for the x values from 4 to 2 and again from 2 to 4. So our answer is [ 4, 2) (2, 4]. 8. To find where f(x) = 1, we look for points on the graph where the y-coordinate is 1. Even though these points aren t specified, we see that the curve has two points with a y value of 1, as seen in the graph below. That means there are two solutions to f(x) = y x As we move from left to right, the graph rises from ( 4, 3) to (0, 3). This means f is increasing on the interval [ 4, 0]. (Remember, the answer here is an interval on the x-axis.) 10. As we move from left to right, the graph falls from (0, 3) to (4, 3). This means f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) 11. The function has its only local maximum at (0, 3). 12. There are no local minimums. Why don t ( 4, 3) and (4, 3) count? Let s consider the point ( 4, 3) for a moment. Recall that, in the definition of local minimum, there needs to be an open interval I which contains x = 4 such that f( 4) < f(x) for all x in I different from 4. But if we put an open interval around x = 4 a portion of that interval will lie outside of the domain of f. Because we are unable to fulfill the requirements of the definition

12 1.7 Graphs of Functions 75 for a local minimum, we cannot claim that f has one at ( 4, 3). The point (4, 3) fails for the same reason no open interval around x = 4 stays within the domain of f. 13. The maximum value of f is the largest y-coordinate which is The minimum value of f is the smallest y-coordinate which is The graph appears to be symmetric about the y-axis. This suggests 13 that f is even. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to find the local maximums and local minimums analytically; this is the business of Calculus. 14 When we have need to find such beasts, we will resort to the calculator. Most graphing calculators have Minimum and Maximum features which can be used to approximate these values, as demonstrated below. Example Let f(x) = 15x x 2. Use a graphing calculator to approximate the intervals on + 3 which f is increasing and those on which it is decreasing. Approximate all extrema. Solution. Entering this function into the calculator gives. Using the Minimum and Maximum features, we get 13 but does not prove 14 Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra.

13 76 Relations and Functions To two decimal places, f appears to have its only local minimum at ( 1.73, 4.33) and its only local maximum at (1, 73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn t be too surprising. The function appears to be increasing on [ 1.73, 1.73] and decreasing on (, 1.73] [1.73, ). This makes 4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Example Find the points on the graph of y = (x 3) 2 which are closest to the origin. Round your answers to two decimal places. Solution. Suppose a point (x, y) is on the graph of y = (x 3) 2. Its distance to the origin, (0, 0), is given by d = (x 0) 2 + (y 0) 2 = x 2 + y 2 = x 2 + [(x 3) 2 ] 2 Since y = (x 3) 2 = x 2 + (x 3) 4 Given a value for x, the formula d = x 2 + (x 3) 4 is the distance from (0, 0) to the point (x, y) on the curve y = (x 3) 2. What we have defined, then, is a function d(x) which we wish to minimize over all values of x. To accomplish this task analytically would require Calculus so as we ve mentioned before, we can use a graphing calculator to find an approximate solution. Using the calculator, we enter the function d(x) as shown below and graph. Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x = To find the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation to get y = (x 3) 2 = (2.00 3) 2 = So, our final answer is (2.00, 1.00). 15 (What does the y value listed on the calculator screen mean in this problem?) 15 It seems silly to list a final answer as (2.00, 1.00). Indeed, Calculus confirms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the author is a pedant, and as such, uses the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such.

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