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 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 Identifying the Graphs of Polynomial Functions 10.8 Review Questions 10.9 Explanations
Explanations

1.      B

This question tests your understanding of compound functions. If f(x) = ax and g(x) = bx, f(g(x)) = f(bx) = abx. From this point, the question is all algebra to solve the equation abx = 5x2 + 2. By factoring an x out of the right side, you get (ab)x = x(5x + 2/x). Divide both sides of this equation by x and you see that ab = 5x + 2/x.

2.      C

To evaluate this compound function, you must first find the inverses of f and g. The inverse is found by interchanging the places of x and y and solving for y: f–1(x) = x/ 2, and g–1(x) = . Thus, f–1(g–1(x)) = / 2. Substituting 2 for x into this equation, the result is /2 = 2/2 = 1.

3.      C

In order to solve this problem, it is simpler to rearrange the problem so it looks like h(x) = (g(f(3)) – 1. Then just substitute the expression for h(x) into the equation:

Remember that you were given h(x) = 3x, and now you know that h(x) = 3. Equate the two and you get 3x = 3 and x = 1.

4.      C

By graphing the function on your calculator, you should be able to see that statements I and II must be true and statement III is false, making C the correct answer choice. Even without a calculator, though, you can calculate the asymptotes of this graph.

A vertical asymptote occurs in this graph where the function is undefined. The function is undefined at x = 2, so this is a vertical asymptote. By plugging in a series of numbers, you find that values of the function approach 2 but are never equal to 2. So there is a horizontal asymptote, the line y = 2.

One easy way to check your asymptotes is to plug the corresponding x and y values into the function. For example, when you plug in x = 2, the function should be undefined. To be sure that x = 2 is indeed an asymptote and not a hole, plug in values on both sides of x = 2, like x = ± 2.01, and make sure that one is very large and one is very small. This indicates that on one side of x = 2, the function approaches infinity, and on the other side the function approaches negative infinity.

5.      B

All of the answer choices are polynomials. You can analyze the polynomial’s end behavior to narrow the choices. By observing the end behavior of f, you can determine whether the leading coefficient is positive or negative, and you can tell whether the degree of f is odd or even. From the graph, you can see that f(x) increases without bound as x increases and as x decreases. This means that the degree of f is even, and the leading coefficient is positive. This eliminates C and D as possible answers.

To choose between the remaining three possible answers requires closer analysis. First, note from the figure that x = 0 is not a root of the function, but x = 0 is a root of the function in A. Therefore, this choice can be eliminated. Second, in the graph, f takes on positive and negative values. The function in E is positive for all values of x, because an even power of x is always positive, and c2 is always positive. So, E can be eliminated from the answer choices, making B the only remaining possibility.

This is a difficult question, and the answer probably didn’t jump out at you right away. Remember that when you have to analyze the graph of a function or a function itself, you can use end behavior, roots, and which portions of the graph have positive and negative values of f as tools for your analysis.

6.      E

The condition that f(x) = –f(–x) means that f is an odd function, which means that it is symmetrical with respect to the origin. The easiest way to answer this question is to choose which of the functions graphed are symmetrical with respect to the origin. The only graphs that satisfy this condition are the graphs in D and E. Note, however, that the graph in D is actually not even the graph of a function, because it doesn’t pass the vertical line test. The correct answer must be E.

7.      E

To find the maximum value of the function, first multiply it out.

The graph of the function f(x) = –2x2 + 28 is a parabola opening downward, since the coefficient of x2 is negative. This means the maximum value of the function will be the y value of the vertex. To find the vertex of the parabola, set x = 0 and solve for f(x). For this function, f(0) = –2(0)2 + 28 = 0 + 28 = 28, so the vertex of the parabola is the point (0, 28), and the maximum value of f is 28.

8.      D

In order for x must equal x – y and y must equal x + y. These equations can be solved simultaneously:

When x = y, the requirement that is fulfilled for this function.

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