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Unformatted text preview: her, the result is always an integer.) Choice (A), which incorporates the concept of absolute value, cannot be the correct answer, since the absolute value of any integer is by definition a positive integer. (E) Either x – 3 > 4 or x – 3 < –4. Solve for x in both inequalities: x > 7; x < –1. (B) If x = 0, y = –1. The point (0,–1) on the graph shows this functional pair. For all negative values of x, y is the absolute value of x, minus 1 (the graph is translated down one unit). The portion of the graph to the left of the yaxis could show these values. For all positive values of x, y = x, minus 1 (the graph is translated down one unit). The portion of the graph to the right of the yaxis could show these values. (D) Substitute 2 for x in the function:
f
1 2. 3. 4. (D) The expression given in the question is equivalent to 4 · 4n. In this expression, base numbers are the same. Since the terms are multiplied together, you can combine exponents by adding them together: 4 · 4n = 4(n+1). (A) Raise both the coefficient –2 and variable x2 to the power of 4. When raising an exponent to a power, multiply together the exponents: (–2x2)4 = (–2)4x(2)(4) = 16x8 3. 4. (C) Any term to a negative power is the same as “one over” the term, but raised to the positive power. Also, a negative number raised to a power is negative if the exponent is odd, yet positive if the exponent is even: –1(–3) + [–1(–2)] + [–12] + [–13] = − + + 1 – 1 11 =0
11 5. ( )= 1 −3 −
1 2 1 2 1 2 1 5. = 2−3 − 2 = −1 − 2
1 The correct answer is 16. Express fractional exponents as roots, calculate the value of each term, and then add:
4 3 2 + 4 3 2 = 4 3 + 4 3 = 64 + 64 = 8 + 8 = 16 = 1− 2
1 = 1 2 www.petersons.com 268 Chapter 15 Exercise 5
1. (A) One way to approach this problem is to substitute each answer choice for x in the function, then find f(x). Only choice (A) provides a value for which f(x) = x:
f Exercise 6
1. (B) To determine the function’s range, apply the rul...
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 Spring '10
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