SAT SUBJECT TEST MATH LEVEL 2

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 1
Functions

 
1.6 Miscellaneous Functions

PIECEWISE FUNCTIONS

Piecewise functions are defined by different equations on different parts of their domain. These functions are useful in modeling behavior that exhibits more than one pattern.

EXAMPLES

1. Graph the function

You can graph this on your graphing calculator by using the 2nd TEST command to enter the symbols < and ≥. Enter (3 – x2)(< 1) + (x3 – 4x)(≥ 1) into Y1. For values of less than 1, (– 1) = 1 and (≥ 1) = 0, so for these values only 3 – x2 will be graphed. The reverse is true for values of greater than 1, so only x3 – 4will be graphed. This graph is shown on the standard grid in the figure below.

Absolute value functions are a special type of piecewise functions. The absolute value function is defined as 

The general absolute value function has the form f(x) = a|– h| + k, with the vertex at (h,k) and shaped like v if > 0 and like ^ if < 0. The vertex separates the two branches of the graph; delineates the domain of all real numbers into two parts. The magnitude of determines how spread out the two branches are. Larger values of correspond to graphs that are more spread out.

The absolute value command is in the MATH/NUM menu of your graphing calculator. You can readily solve absolute value equations or inequalities by finding points of intersection.

2. If |– 3| = 2, find x.

Enter |– 3| into Y1 and 2 into Y2. As seen in the figure below, the points of intersection are at = 5 and = 1.

This is also easy to see algebraically. If |– 3| = 2, then – 3 = 2 or – 3 = –2. Solving these equations yields the same solutions: 5 or 1. This equation also has a coordinate geometry solution: |– b| is the distance between and b. Thus |– 3| = 2 has the interpretation that is 2 units from 3. Therefore, must be 5 or 1.

3. Find all values of for which |2+ 3| ≥ 5.

The graphical solution is shown below.

The desired values of are on, or right and left of, the points of intersection: ≥ 1 or  –4. By writing the inequality as , we can also interpret the solutions to the inequality as those points that are more than  units from .

4. If the graph of f(x) is shown below, sketch the graph of (A) | f(x)| (B) f(|x|).

(A) Since |f(x)| ≥ 0, by the definition of absolute value, the graph cannot have any points below the x-axis. If f(x) < 0, then |f(x)| = –f(x). Thus, all points below the x-axis are reflected about the x-axis, and all points above the x-axis remain unchanged.

(B) Since the absolute value of is taken before the function value is found, and since |x| = –when < 0, any negative value of will graph the same y-values as the corresponding positive values of x. Thus, the graph to the left of the y-axis will be a reflection of the graph to the right of they-axis.

5. If f(x) = |+ 1| – 1, what is the minimum value of f(x)?

Since |+ 1| ≥ 0, its smallest value is 0. Therefore, the smallest value of f(x) is 0 – 1 = –1. The graph of f(x) is indicated below.

Step functions are another special type of piecewise function. These functions are constant over different parts of their domains so that their graphs consist of horizontal segments. The greatest integer function, denoted by [x], is an example of a step function. If is an integer, then [x] = x. If is not an integer, then [x] is the largest integer less than x.

The greatest integer function is in the MATH/NUM menu as int on TI-83/84 calculators.

6. Five examples of the greatest integer function integer notation are:

           (1) [3.2] = 3

           (2) [1.999] = 1

           (3) [5] = 5

           (4) [–3.12] = –4

           (5) [–0.123] = – 1.

7. Sketch the graph of f(x) = [x].

TIP 

On the graphing calculator, you can’t tell which side of each horizontal segment has the open and closed point.

8. What is the range of .

Enter int(int(x)/x) into Y1 and choose Auto for both Indpnt and Depend in TBLSET. Set TblStart to 0 and Tbl to 0.1. Inspection of TABLE shows only 0s and 1s as Y1, so the range is the two-point set {0,1}.

EXERCISES

1.       |2– 1| = 4+ 5 has how many numbers in its solution set?

           (A)  0

           (B)  1

           (C)  2

           (D)  an infinite number

           (E)  none of the above

2.       Which of the following is equivalent to 1  |– 2|  4?

           (A)  3   6

           (B)   1 or ≥ 3

           (C)  1   3

           (D)   –2 or ≥ 6

           (E)  –2   1 or 3   6

3.       The area bound by the relation |x| + |y| = 2 is

           (A)  8

           (B)  1

           (C)  2

           (D)  4

           (E)  There is no finite area.

4.       Given a function, f(x), such that f(x) = f(|x|). Which one of the following could be the graph of f(x)?

           (A)  

           (B)  

           (C)  

           (D)  

           (E)  

5.       The figure shows the graph of which one of the following?

           (A)  = 2– |x|

           (B)  = |– 1| + x

           (C)  = |2– 1|

           (D)  = |+ 1| – x

           (E)  = 2|x| – |x|

6.       The postal rate for first-class mail is 44 cents for the first ounce or portion thereof and 17 cents for each additional ounce or portion thereof up to 3.5 ounces. The cost of a 3.5-ounce letter is 95¢. A formula for the cost in cents of first-class postage for a letter weighing ounces (3.5) is

           (A)  44 + [– 1] · 17

           (B)  [– 44] · 17

           (C)  44 + [N] · 17

           (D)  1 + [N] · 17

           (E)  none of the above

7.       If f(x) = i, where is an integer such that  + 1, the range of f(x) is

           (A)  the set of all real numbers

           (B)  the set of all positive integers

           (C)  the set of all integers

           (D)  the set of all negative integers

           (E)  the set of all nonnegative real numbers

8.       If f(x) = [2x] – 4with domain 0   2, then f(x) can also be written as

           (A)  2x

           (B)  –x

           (C)  –2x

           (D)  x2 – 4x

           (E)  none of the above