Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
 10.1 Characteristics of a Function 10.2 Evaluating Functions 10.3 Compound Functions 10.4 Inverse Functions

 10.5 Domain and Range 10.6 Graphing Functions 10.7 Review Questions 10.8 Explanations
Explanations

1.      A

A function is an association between two sets such that each member of the first set, the domain, is paired with exactly one member of the second set, the range. In order for one of the sets above to be a function, the same input (x-coordinate) cannot be assigned to more than one output (y-coordinate). This is true of all the choices except the first, in which the input 0 is assigned to three different outputs. A is not a function.

2.      D

Plug in 3 for s, and then solve for t0:

Since it is given that f(3, t0) = 0, we can set (t06)2 = 0. Thus, t0 = 6.

3.      C

The function f /g(x) can be simplified to f(x) /g(x). Since g(x) is the denominator, the function f /g(x) is undefined for values at which g(x) = 0. From the given information, we know these values are 3 and 1. To find which function could be g(x), simply test each answer choice to see if plugging 3 and 1 into the function produces zero.

4.      E

To find the inverse of a function, set the function equal to y, interchange the places of y and x, and solve for y. The result is the inverse of your function:

5.      D

Assume the domain of the function is all real numbers. To find the points at which the function is undefined, restrict it such that the quantity under the square root (in the numerator) is greater than or equal to zero and that the denominator is not equal to zero. The quantity under the square root, x + 3, is greater than or equal to zero if x ≥ –3. The denominator can be simplified to x3 – 2x2 – 8x = x(x2 – 2x – 8) = x(x + 2)(x – 4). Thus, the values of x that make the denominator equal to zero are x = {–2, 0, 4}. This means that the domain of the function is the real numbers x such that x ≥ –3, x ≠ {–2, 0, 4}.

6.      D

Since we’re given the range for x+5/2, we can use algebra to find the range of |x|:

At this point, you can write two equations: –3 < x < –1 and 1 < x < 3. From these equations, you know 1 < |x| < 3. Thus, 2 is the only possible correct answer choice.

7.      E

Use the vertical line test: the only graph that a vertical line will not cross twice is the last answer choice, E.

8.      E

The range is the set of all y-values the graph touches, and the domain is the set of all x-values for which the graph is defined. According to these definitions, you can simply look at the graph and see that the range is all real numbers less than or equal to 4, y ≤ 4, and that the domain is the set of real numbers not equal to zero, because at x = 0, there is a vertical asymptote. Therefore, the last answer choice is correct.

9.      C

The number of x-intercepts equals the number of roots. Calculate the roots of f(x) by setting the function equal to 0. When f(x) = 0, x has four possible values: 0, 4, –3 and –7. These values are the roots of f(x), so f(x) has 4 roots, or 4 x-intercepts.

You can use a graphing calculator to check this answer. Graph f(x) and see how many times f(x) intersects the x-axis.

10.      D

2u + 3v = 2(2, y) + 3(x, –4) = (4, 2y) + (3x, –12) = (4 + 3x, 2y – 12). Set this ordered pair equal to (10, –12), and simple algebra shows x = 2 and y = 0.

 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
Test Prep Centers
SparkCollege
 College Admissions Financial Aid College Life