Bluebook SAT Math: Advanced Functions Questions and Answers 2

Question 1

The function f(x) is an even function and the function g(x) is an odd function. Which of the following describes the function h(x) = 2f(x) - f(x)g(x)?

Accepted Answer

Neither even nor odd

Answer

To determine if h(x) is even, odd, or neither, we must evaluate h(-x). By definition, f is even, so f(-x) = f(x). By definition, g is odd, so g(-x) = -g(x). Let's substitute -x into the function h(x): h(-x) = 2f(-x) - f(-x)g(-x). Now, substitute the properties of even and odd functions: h(-x) = 2f(x) - [f(x)][-g(x)]. This simplifies to h(-x) = 2f(x) + f(x)g(x). Now, compare h(-x) with h(x) and -h(x). h(x) = 2f(x) - f(x)g(x). -h(x) = -2f(x) + f(x)g(x). Since h(-x) is not equal to h(x) and is not equal to -h(x), the function h(x) is neither even nor odd.

Question 2

The function f(x) = x^3 - x^2 - 4x + 4 has three distinct real roots. A new function is defined as g(x) = |f(x)|. How many local minima does the graph of y = g(x) have?

Accepted Answer

Three

Answer

First, find the roots of f(x) by factoring: f(x) = x^2(x - 1) - 4(x - 1) = (x^2 - 4)(x - 1) = (x - 2)(x + 2)(x - 1). The roots are x = -2, 1, and 2. The graph of f(x) is a cubic polynomial that crosses the x-axis at these three points. Between the roots, the function will have one local maximum (between x=-2 and x=1) and one local minimum (between x=1 and x=2). When we take the absolute value to create g(x) = |f(x)|, any part of the graph of f(x) that was below the x-axis is reflected above it. The original local minimum of f(x) was below the x-axis, so its reflection becomes a local maximum for g(x). The original local maximum of f(x) was above the x-axis, so it remains a local maximum for g(x). The key is that the three x-intercepts (roots), which were points where the graph crossed the axis, now become points where the graph touches the axis and turns around. These points become local minima for g(x). Therefore, g(x) has three local minima, located at x = -2, x = 1, and x = 2.

Question 3

Let f(x) = √(x - 3) and g(x) = x^2 + 2. What is the domain of the composite function h(x) = f(g(x))?

Accepted Answer

x ≥ 1 or x ≤ -1

Answer

The composite function is h(x) = f(g(x)). To find its domain, we must ensure that the output of the inner function, g(x), is in the domain of the outer function, f(x). The domain of f(x) = √(x - 3) requires the expression inside the square root to be non-negative, so x - 3 ≥ 0, which means the input to f must be ≥ 3. Therefore, we must find the values of x for which the output of g(x) is ≥ 3. We set up the inequality: g(x) ≥ 3, which means x^2 + 2 ≥ 3. Subtracting 2 from both sides gives x^2 ≥ 1. This inequality is true when x ≥ 1 or when x ≤ -1. Thus, the domain of h(x) is (-∞, -1] U [1, ∞).

Question 4

A rational function is given by f(x) = (ax^2 + 5x - 2) / (3x^2 - bx - 12). If the function has a vertical asymptote at x = -3/2 and a horizontal asymptote at y = 2, what is the value of b?

Accepted Answer

-4

Answer

First, use the horizontal asymptote to find 'a'. For a rational function where the degree of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. So, y = 4/a. We are given the asymptote is y=2, so 4/a = 2, which means a=2. Next, use the vertical asymptote to find 'b'. A vertical asymptote occurs where the denominator equals zero. We know the denominator is 2x^2 + bx - 6, and it equals zero at x=3. Substitute x=3 into the denominator: 2(3)^2 + b(3) - 6 = 0. This simplifies to 2(9) + 3b - 6 = 0, which is 18 + 3b - 6 = 0. Combining terms gives 12 + 3b = 0. Solving for b, we get 3b = -12, so b = -4.

Question 5

A polynomial P(x) has real coefficients. The graph of y=P(x) has x-intercepts at x=2 and x=-1. If 3 - i is also a root of P(x)=0, what is the lowest possible degree of P(x)?

Accepted Answer

4

Answer

The degree of a polynomial is determined by its number of roots. We are given three roots: 2, -1, and 3 - i. The Complex Conjugate Root Theorem states that if a polynomial has real coefficients, then its complex roots must come in conjugate pairs. Since 3 - i is a root, its complex conjugate, 3 + i, must also be a root. Therefore, the polynomial must have at least four roots: 2, -1, (3 - i), and (3 + i). A polynomial with four roots must have a degree of at least 4. This is the lowest possible degree.

Bluebook SAT Practice Test Practice Test

Bluebook SAT Practice Test Practice Test

Bluebook SAT Math: Advanced Functions Questions and Answers 2