Domain & Range - Accuplacer Aaf

1. What is the range of the function $g(x) = -2|x - 3| + 4$?
   Answer: $(-\infty, 4]$ The vertex of the absolute value function is at $(3, 4)$. Because the coefficient -2 is negative, the V-shape opens downward. This means the vertex is a maximum point. The maximum value of the function is 4. Therefore, the range includes all real numbers less than or equal to 4, which is $(-\infty, 4]$.
2. What is the domain of the function $g(x) = \sqrt{x^2 - x - 12}$?
   Answer: $(-\infty, -3] \cup [4, \infty)$ The domain requires the radicand to be non-negative: $x^2 - x - 12 \geq 0$. Factoring the quadratic gives $(x-4)(x+3) \geq 0$. The roots are at $x=4$ and $x=-3$. Since the parabola opens upward, the expression is non-negative at or outside the roots. Therefore, the domain is $x \leq -3$ or $x \geq 4$, which is $(-\infty, -3] \cup [4, \infty)$ in interval notation.
3. What is the domain of the function $g(x) = \sqrt{7 - 2x}$?
   Answer: $x \leq \frac{7}{2}$ The domain of a square root function requires the expression under the radical (the radicand) to be non-negative. Therefore, we must have $7 - 2x \geq 0$. Subtracting 7 from both sides gives $-2x \geq -7$. When dividing both sides by -2, we must reverse the inequality sign, which gives $x \leq \frac{7}{2}$.
4. Consider the function $p(x) = \frac{1}{x^2 + x - 6}$. After factoring the denominator and identifying all restrictions, what is the domain of this function?
   Answer: $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$ When you encounter a rational function (a fraction with a polynomial in the denominator), finding the domain means identifying where the function is undefined. Since division by zero is undefined, you need to find all values of x that make the denominator equal to zero. Start by factoring the denominator: $x^2 + x - 6$. You need two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2, so $x^2 + x - 6 = (x + 3)(x - 2)$. Now set each factor equal to zero: $x + 3 = 0$ gives $x = -3$, and $x - 2 = 0$ gives $x = 2$. These are the restrictions—values where the function is undefined. The domain includes all real numbers except -3 and 2, written as $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$, which is answer choice C. Let's examine why the other answers are wrong. Answer A, $(-\infty, 2) \cup (2, \infty)$, only excludes x = 2 but misses the restriction at x = -3. Answer B, $(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$, incorrectly excludes x = -2 and x = 3—these come from incorrectly factoring or setting up the equation. Answer D, $(-\infty, 3) \cup (3, \infty)$, only excludes x = 3, which isn't even a restriction. Study tip: Always factor the denominator completely and set each factor equal to zero separately. Double-check your factoring by expanding back to the original expression—this prevents careless errors that lead to wrong restrictions.
5. A function $m(x) = \log_2(4 - x^2)$ is defined when the argument of the logarithm is positive. If this function achieves its maximum value on its domain, what is that maximum value?
   Answer: $2$ When you encounter a logarithmic function with restrictions, you need to first identify the domain, then find where the function reaches its peak within that domain. For $m(x) = \log_2(4 - x^2)$ to be defined, the argument must be positive: $4 - x^2 > 0$. Solving this inequality gives us $x^2 < 4$, so $-2 < x < 2$. The domain is the open interval $(-2, 2)$. To find the maximum, recognize that $\log_2$ is an increasing function, so $m(x)$ is maximized when its argument $4 - x^2$ is maximized. Since $4 - x^2$ is a downward-opening parabola with vertex at $x = 0$, the maximum occurs at $x = 0$. At this point, $m(0) = \log_2(4 - 0^2) = \log_2(4) = \log_2(2^2) = 2$. Choice A is wrong because the function does achieve a maximum at $x = 0$ within its domain. Choice B incorrectly gives the value of the argument ($4$) rather than the logarithm itself. Choice C might result from incorrectly calculating $\log_2(4)$ as $1$, perhaps confusing it with $\log_4(4)$. The correct answer is D: $2$. Strategy tip: For logarithmic optimization problems, remember that since logarithmic functions are monotonic (always increasing or decreasing), maximizing the logarithm is equivalent to maximizing its argument. Always check the domain restrictions first, then optimize the expression inside the logarithm.
6. What is the range of the function $f(x) = -x^2 + 6x - 5$?
   Answer: $(-\infty, 4]$ The graph of this quadratic function is a parabola that opens downward because the leading coefficient is negative (-1). The maximum value of the function occurs at the vertex. The x-coordinate of the vertex is $x = -b/(2a) = -6/(2(-1)) = 3$. The y-coordinate (the maximum value) is $f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4$. Thus, the range is all real numbers less than or equal to 4, which is $(-\infty, 4]$.
7. The function $k(x) = |x - 2| - |x + 1|$ can be analyzed by considering the critical points where the expressions inside the absolute values equal zero. What is the range of this function?
   Answer: $[-3, 3]$ Critical points are $x = 2$ and $x = -1$. For $x < -1$: both expressions are negative inside absolute values, so $k(x) = -(x-2) - (-(x+1)) = -x + 2 + x + 1 = 3$. For $-1 \leq x < 2$: $k(x) = -(x-2) - (x+1) = -x + 2 - x - 1 = -2x + 1$, ranging from $k(-1) = 3$ to $k(2) = -3$. For $x \geq 2$: $k(x) = (x-2) - (x+1) = x - 2 - x - 1 = -3$. The function equals 3 for $x < -1$, decreases linearly from 3 to -3 on $[-1, 2]$, and equals -3 for $x \geq 2$. Therefore, range is $[-3, 3]$. Choice B misses the lower bound. Choice C misses the upper bound. Choice D incorrectly assumes only discrete values are achieved.
8. The area $A$ of a circle is given by the function $A(r) = \pi r^2$, where $r$ is the radius. The circle is expanding, and its radius must be greater than 2 cm but no more than 10 cm. What is the domain of this function in this context?
   Answer: $(2, 10]$ The domain represents the set of all possible input values, which in this context is the radius $r$. The problem states that the radius must be 'greater than 2 cm' ($r > 2$) and 'no more than 10 cm' ($r \leq 10$). Combining these two conditions gives $2 < r \leq 10$, which is represented by the interval $(2, 10]$.
9. What is the domain of the function $f(x) = \frac{x+5}{x^2 - 4}$?
   Answer: All real numbers except -2 and 2 The domain of a rational function includes all real numbers except those that make the denominator zero. To find the excluded values, set the denominator equal to zero: $x^2 - 4 = 0$. Factoring gives $(x-2)(x+2) = 0$, so $x=2$ and $x=-2$ are excluded. Therefore, the domain is all real numbers except -2 and 2.
10. Which of the following represents the domain of the function $h(x) = \frac{x}{\sqrt{x-3}}$?
    Answer: $x > 3$ The function has two conditions for its domain. First, the expression under the square root, $x-3$, must be non-negative ($x-3 \geq 0$). Second, the denominator, $\sqrt{x-3}$, cannot be zero. Combining these, the expression $x-3$ must be strictly greater than zero. So, $x-3 > 0$, which means $x > 3$.
11. What is the domain of the function $f(x) = \ln(3x + 9)$?
    Answer: $x > -3$ The domain of a logarithmic function requires its argument to be strictly positive. Therefore, we must have $3x + 9 > 0$. Subtracting 9 from both sides gives $3x > -9$. Dividing by 3 gives $x > -3$.
12. What is the range of the function $f(x) = \frac{6x - 5}{2x + 1}$?
    Answer: All real numbers except $y = 3$ The range of a rational function of the form $\frac{ax+b}{cx+d}$ is all real numbers except the value of its horizontal asymptote. The horizontal asymptote is found by taking the ratio of the leading coefficients of the numerator and the denominator. In this case, the asymptote is at $y = \frac{6}{2} = 3$. Therefore, the range is all real numbers except 3.
13. Let $f(x) = \sqrt{x-2}$ and $g(x) = \frac{1}{x-5}$. What is the domain of the function $(f+g)(x)$?
    Answer: $[2, 5) \cup (5, \infty)$ The domain of a sum of functions, $(f+g)(x)$, is the intersection of the domains of $f(x)$ and $g(x)$. The domain of $f(x) = \sqrt{x-2}$ is $x-2 \geq 0$, or $x \geq 2$. The domain of $g(x) = \frac{1}{x-5}$ is $x-5 \neq 0$, or $x \neq 5$. The intersection of these two sets is all numbers greater than or equal to 2, except for 5. In interval notation, this is $[2, 5) \cup (5, \infty)$.
14. A projectile is fired from the ground. Its height $h$ in meters after $t$ seconds is modeled by the function $h(t) = -5t^2 + 40t$. The projectile hits the ground after 8 seconds. What is the range of this function for the duration of the flight?
    Answer: $[0, 80]$ The range is the set of all possible heights. The minimum height is 0 (at $t=0$ and $t=8$). The maximum height occurs at the vertex of the parabola. The time of the vertex is $t = -b/(2a) = -40/(2(-5)) = 4$ seconds. The maximum height is $h(4) = -5(4)^2 + 40(4) = -80 + 160 = 80$ meters. Therefore, during its flight, the projectile's height ranges from 0 to 80 meters, inclusive. The range is $[0, 80]$.
15. For the function $f(x) = \sqrt{x-3} + 1$, which value is in the range but not in the domain?
    Answer: 2 First, determine the domain and range. The domain requires $x-3 \geq 0$, so $x \geq 3$. The domain is $[3, \infty)$. The range of $\sqrt{x-3}$ is $[0, \infty)$, so the range of $\sqrt{x-3} + 1$ is $[1, \infty)$. We need a number that is in $[1, \infty)$ but not in $[3, \infty)$. Looking at the options: A) 0 is in neither. B) 2 is in the range (since $2 \geq 1$) but not in the domain (since $2 < 3$). C) 4 is in both. D) 10 is in both. Thus, 2 is the correct answer.
16. The domain of a function $f(x)$ is $[-4, 6]$. What is the domain of the function $g(x) = f(2x)$?
    Answer: $[-2, 3]$ The domain of $f(x)$ specifies the allowed values for its input. In the function $g(x) = f(2x)$, the input to $f$ is $2x$. Therefore, $2x$ must be within the domain of $f$. This gives the inequality $-4 \leq 2x \leq 6$. To find the domain for $x$, we divide all parts of the inequality by 2, resulting in $-2 \leq x \leq 3$. The domain of $g(x)$ is $[-2, 3]$.
17. What is the domain of the function $h(x) = \frac{x+1}{\sqrt{x^2 - 9}}$?
    Answer: $(-\infty, -3) \cup (3, \infty)$ The expression under the square root in the denominator must be strictly positive (it cannot be negative because of the root, and it cannot be zero because it's in the denominator). So we must solve the inequality $x^2 - 9 > 0$. This is equivalent to $x^2 > 9$. Taking the square root of both sides gives $|x| > 3$, which means $x > 3$ or $x < -3$. In interval notation, the domain is $(-\infty, -3) \cup (3, \infty)$.
18. What is the domain of the function $f(x) = \frac{3x}{x^2 + 2x + 5}$?
    Answer: All real numbers The domain of a rational function is all real numbers except for values of $x$ that make the denominator zero. We check if $x^2 + 2x + 5$ can be zero by calculating its discriminant: $b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$. Since the discriminant is negative, the quadratic equation $x^2 + 2x + 5 = 0$ has no real solutions. The denominator is never zero, so there are no restrictions on the domain. The domain is all real numbers.
19. The function $g(x) = \sqrt{16 - x^2} + 3$ has domain restrictions due to the square root. After determining the domain, what is the complete range of this function?
    Answer: $[3, 7]$ The domain requires $16 - x^2 \geq 0$, so $x^2 \leq 16$, giving domain $[-4, 4]$. The expression $\sqrt{16 - x^2}$ represents the upper half of a circle with radius 4, so its range is $[0, 4]$. Adding 3 shifts this up to $[3, 7]$. The minimum occurs at the endpoints $x = ±4$ where $g(±4) = \sqrt{0} + 3 = 3$, and the maximum occurs at $x = 0$ where $g(0) = \sqrt{16} + 3 = 7$. Choice B gives the domain instead of range. Choice C ignores the vertical shift of +3. Choice D incorrectly assumes the function increases without bound.
20. The range of a function $f(x)$ is $[-2, 10]$. What is the range of the function $g(x) = 3f(x) - 1$?
    Answer: $[-7, 29]$ The range of $f(x)$ is the set of its output values. The function $g(x)$ takes the output of $f(x)$, multiplies it by 3, and then subtracts 1. We apply these transformations to the endpoints of the range of $f$. The lower bound becomes $3(-2) - 1 = -6 - 1 = -7$. The upper bound becomes $3(10) - 1 = 30 - 1 = 29$. Therefore, the range of $g(x)$ is $[-7, 29]$.

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