Solution for Explore! on Page 5

For Y1(1) we obtain a nonreal answer, because the square root of -1 is the imaginary number i. This result shows you how important it is to choose values of X that have well-defined functional values.

When you evaluate Y2(1), you should obtain Y2(1) = an undefined value (division by zero) and Y2(2) = 1. These results differ from that of the function in Example 1.3, which has been defined in pieces, called a piecewise function. Over the interval x ³ 1, f(x) assumes the rule 3x² + 1. This rule shows you the importance of knowing the appropriate formula to apply to each x interval of values. Hence, f(2) = 13, whereas Y2(2) = 1.

Note that the functions can be input into the function editor (Y= key) either as separate functions or defined in terms of the original function. To define Y2 in terms of Y1, press the VARS key, then the right arrow to select Y-VARS. Press 1 for Function, then 1 for Y1.

After you have entered the three functions into the equations editor, press graph to obtain the following graphical solutions:

The domain of g(x) is X ³ 2, while that of h(x) is X ³ -4.

The composites f(g(x)) and f(h(x)) can be placed into the function editor as Y1(Y2) and Y1(Y3), respectively. You can compare the graph of f(x) and f(g(x)) by deselecting the second, third and fifth functions. To deselect a function, press the Y= key. Using the arrow keys, move the cursor to the function’s = sign. Press enter to deselect the function. Selected functions should have a highlighted = sign.

Notice that f(g(x)) is just the square function with its vertex translated over to x = -3.

Notice that the composite f(h(x)), graphed in bold, is not the function y = x over the entire real line but restricted to the interval x ³ 0. The domain of the composite f(h(x)) is at most the domain of h(x).

Create the composite functions Y2(Y1) and Y3(Y1). Compare Y2(Y1) with Y1(Y2). You observe that the first composite is y = x² + 3, while the second is y = (x + 3)²

You can observe similar results for Y3(Y1) and Y3(Y1) by selective graphing.

Observe that h(f(x)), namely Y3(Y1), is the absolute value function, abs(x).

Place the functions into the function editor in the following manner and graph:

You can see that Y2 and Y3 are vertical translations of Y1, because a constant is added to the function.

On the other hand, compare Y1 with Y3 and Y4. Changing the argument x by (x + 2) and

(x - 3) results in horizontal translations of the original square function. What are simple expression for Y3 and Y4?

Be careful to choose the correct graphing mode for certain examples. You will obtain the representation below on the left, unless you change the graphing mode to DOT instead of CONNECTED. Press the MODE key and use the arrow keys to select Dot mode. You will see that the graph on the right is more accurate.

The function in Example 2.2 is written as one expression using indicator functions.

The TRACE key has been used to locate the point (4,3).

Using the Decimal Window below insures decimal values for use of the TRACE key. You can switch to decimal window by pressing the ZOOM key and selecting the 4:ZDecimal option.

Using a decimal window and the TRACE key, you can locate one x-intercept as x = -1. Using the 2:zero option of the CALC (2nd TRACE) menu, specify the left bound, the right bound and an estimate for the calculator to determine the root. Use x = 2 for the one on the right.

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For g(x) = x² + x - 4, we obtain the two values below:

Note the intercepts for the three graphs:

Graphing the cost function Y1 = 50x + {200, 300, 400} reveals a series of parallel cost functions. All the functions have the same slope but differing y-intercepts. Check the varying costs for x = 3 units. Don’t forget to set your viewing window to the specified parameters.

Let L1 = {4, 1, 0, -1, .3}.

Let L1 = {-1.5, -.5, 0,. 5, 2.5}.

Using a square viewing window, for A = -1 (-1 ® A) and you obtain the following perpendicular lines:

The exact point of intersection is (-1.5, 3.5). For A > 0, the point of intersection is in the first quadrant; for A < 0 this point is in the second quadrant. What happens when A = 0? Note that if a standard window is used, the lines do not appear perpendicular.

Using D Tbl = 100 and 50 are not as helpful as the value 10, for which the table shows a possible minimal value in the range between x = 80 to 120 with corresponding functional values around 200.

Hence, using values in these ranges allows you to construct a viewing window displaying the graph below:

Using D Tbl = 1, yields the following table and reasonable window values:

Using the TRACE key, you can get close to a minimum value, or you can use the minimum finding feature through the CALC (2nd TRACE) menu, although the answer may still not be exact.

For a cost of $3.00 (C = 300), there appears to be two radial answers. For C = $2.00 there is none.

Exploring values for p via the table feature can be revealing. For instance, setting D Tbl = 5 yields the same functional value of 390 at x = 20. Even if the functional values were close but not exactly the same, we would get a good idea of a reasonable window, say, [15, 25]1 by [300,450]10, to view the intersection point of the two curves:

Using the TBLSET feature (2^nd WINDOW), set D Tbl = .01 and TblStart = .9. We notice that at x = 1, the function is undefined (division by zero), but it appears that the functional values are close to a value of 3.00. Tracing the graph also displays y-values hovering about 3.00.

However, we cannot obtain this f(x) value at the exact point that x = 1. Thus, we say that the function f(x) is undefined at x = 1 but has a limiting value of 3.00 at x = 1.

In this case the limit of f(x) as x approaches 2 does not exist. However, we can write:

lim f(x) = ¥ , recognizing that the y values become unbounded from either side x = 2.

x ® 2

The value of y when x is near 2 depends on the direction from which x is approached. For example, when x is less than 2, y = 2. In fact, when x = 2, y = 3. However, when x is greater than, but not equal to 2, y = 5.

Approaching x = 2 from the left side yields large negative functional values. Approaching x = 2 from the right side, you get large positive values. The function is undefined at x = 2. (A vertical asymptote occurs at x = 2.)

The function y is not defined at x = 1 and a small hole is noticeable on the graph. However, the limit of y as x approaches 1 appears to be 1/2. This can be seen by tracing close to x = 1 or by examining the suggested table of values.

This function is not continuous, because a vertical asymptote at appears x = 2. However, the function is continuous at x = 3, taking on the well-defined value y =4 at that point, as well as approaching this value 4 from both sides as x nears 3.

Although f(1) = 1, the function is not continuous at x = 1. We know this because a break in the graph is evident. Approaching x = 1 from the left side, you obtain a y-value limit of 2. Approaching x = 1 from the right side, you obtain the y-value 1.

The graph appears continuous using a standard window. However, the modified decimal window show an exaggerated hole at x = 2. The function is not continuous, specifically at x = 2 where it is not defined. This case is similar to Example 6.3 (b). What is the value of y that would fill up the hole in the graph?

The value needed to fill up the hole is lim f(x) = 12.

x® 2