Functions & Function Notation - Math 1

1. Which of the following relations represents a function where the domain is $\{-2, -1, 0, 1, 2\}$?
   Answer: $\{(-2, 3), (-1, 2), (0, 1), (1, 2), (2, 3)\}$ A relation is a function if each input appears exactly once. Choice A contains $(-1, 1)$ and $(-1, -1)$, so -1 maps to two different outputs. Choice C contains $(0, 0)$ and $(0, -1)$, so 0 maps to two different outputs. Choice D is missing the input 0 from the specified domain. Only choice B has each domain element appearing exactly once with a unique output.
2. Let $f(x) = \frac{2x + 1}{x - 3}$. If $f(a) = f\left(\frac{1}{a}\right)$ and $a \neq 0, 3, \frac{1}{3}$, what is the value of $a^2$?
   Answer: $1$ Setting $f(a) = f\left(\frac{1}{a}\right)$: $\frac{2a + 1}{a - 3} = \frac{2 \cdot \frac{1}{a} + 1}{\frac{1}{a} - 3} = \frac{\frac{2 + a}{a}}{\frac{1 - 3a}{a}} = \frac{2 + a}{1 - 3a}$. Cross-multiplying: $(2a + 1)(1 - 3a) = (2 + a)(a - 3)$. Expanding the left side: $2a - 6a^2 + 1 - 3a = -6a^2 - a + 1$. Expanding the right side: $2a - 6 + a^2 - 3a = a^2 - a - 6$. Setting them equal: $-6a^2 - a + 1 = a^2 - a - 6$, so $-7a^2 + 7 = 0$, which gives $a^2 = 1$.
3. If $f(x) = 3x - 1$ and $f(a + 2) = f(2a - 1)$, what is the value of $f(a^2)$?
   Answer: $26$ First, find a by using f(a+2) = f(2a-1). Since f(x) = 3x-1, we have f(a+2) = 3(a+2)-1 = 3a+5 and f(2a-1) = 3(2a-1)-1 = 6a-4. Setting them equal: 3a+5 = 6a-4, so 9 = 3a, giving a = 3. Then f(a²) = f(9) = 3(9)-1 = 26. Choice B results from incorrectly calculating f(a²) as f(2a). Choice C comes from finding a² = 9 but forgetting to apply the function. Choice D results from arithmetic errors in solving for a.
4. If $g(x) = 3x - 2$ and $g(g(x)) = 9x - 8$, what is the value of $g^{-1}(4)$?
   Answer: $2$ First, let's verify the given information. If $g(x) = 3x - 2$, then $g(g(x)) = g(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8$ ✓. To find $g^{-1}(4)$, we need to solve $g(x) = 4$: $3x - 2 = 4$, so $3x = 6$, and $x = 2$. Therefore, $g^{-1}(4) = 2$.
5. Consider the relation $R = \{(x, y) : y = \pm\sqrt{x + 3}\}$ for $x \geq -3$. Which statement about this relation is correct?
   Answer: $R$ is not a function because some input values correspond to two output values A relation is a function if and only if each input (x-value) corresponds to exactly one output (y-value). The relation $R = \{(x, y) : y = \pm\sqrt{x + 3}\}$ means that for most x-values in the domain, there are two corresponding y-values: one positive and one negative. For example, when $x = 1$, we get $y = \pm\sqrt{4} = \pm 2$, so both $(1, 2)$ and $(1, -2)$ are in the relation. This violates the definition of a function.
6. The function $p(x)$ satisfies $p(x + 2) = 2p(x) + 3$ for all real numbers $x$. If $p(0) = 1$, what is the value of $p(4)$?
   Answer: $13$ When you encounter a functional equation like $p(x + 2) = 2p(x) + 3$, you're dealing with a recurrence relation. The key is to use the given initial condition and apply the rule step-by-step to find the desired value. Starting with $p(0) = 1$, you can find $p(2)$ by substituting $x = 0$ into the functional equation: $p(0 + 2) = 2p(0) + 3$ $p(2) = 2(1) + 3 = 5$ Next, find $p(4)$ by substituting $x = 2$: $p(2 + 2) = 2p(2) + 3$ $p(4) = 2(5) + 3 = 13$ Looking at the wrong answers: Choice A ($11$) might result from incorrectly calculating $p(4) = 2p(2) + 1$ instead of adding 3. Choice B ($23$) could come from mistakenly using $p(4) = 4p(2) + 3$, perhaps confusing the coefficient with the step size. Choice C ($17$) might arise from computing $p(4) = 2p(2) + 7$, possibly from adding the wrong constant term. The correct answer is D ($13$). For functional equations, always work systematically from your known values. Write out each step clearly, substituting the correct $x$-value into the given rule. Double-check that you're using the exact coefficients and constants from the original equation—small arithmetic errors in the pattern will compound as you iterate forward.
7. Given that $f(x) = x^2 + 2x + 1$ and $h(x) = x + 1$, if $f(x) = h(x) \cdot q(x)$ for some function $q(x)$, what is $q(3)$?
   Answer: $4$ This problem involves polynomial division and the relationship between factors and quotients. When you see an equation like $f(x) = h(x) \cdot q(x)$, you're looking at a factorization where $h(x)$ is a factor of $f(x)$, and $q(x)$ is the quotient. First, let's examine $f(x) = x^2 + 2x + 1$. This is a perfect square trinomial that factors as $(x + 1)^2$. Since $h(x) = x + 1$, we can write: $f(x) = (x + 1)^2 = (x + 1)(x + 1) = h(x) \cdot (x + 1)$ This means $q(x) = x + 1$. Therefore, $q(3) = 3 + 1 = 4$. Let's check why the other answers are incorrect. Choice A) $3$ would be the result if you mistakenly thought $q(x) = x$ and evaluated at $x = 3$. Choice B) $6$ might come from incorrectly adding $f(3) = 16$ and $h(3) = 4$, then subtracting something arbitrarily. Choice C) $5$ could result from thinking $q(x) = x + 2$ and evaluating at $x = 3$. You can verify this is correct by checking: $f(3) = 3^2 + 2(3) + 1 = 16$ and $h(3) \cdot q(3) = 4 \cdot 4 = 16$ ✓ Study tip: When you see polynomial division problems, always look for factoring opportunities first. Perfect square trinomials like $x^2 + 2x + 1$ appear frequently and factor as $(x + 1)^2$. Recognizing common patterns saves time and reduces errors.
8. Consider the relation $R = \{(x, y) : y = \pm\sqrt{x - 3}\}$ where $x \geq 3$. If $g(x) = -\sqrt{x - 3}$ for $x \geq 3$, which statement correctly describes the relationship between $R$ and $g$?
   Answer: $R$ is not a function, but $g$ represents the lower half of $R$ and is itself a function The relation R assigns two y-values (positive and negative square roots) to each x-value in its domain, violating the definition of a function. However, g(x) = -√(x-3) assigns exactly one y-value to each x-value, making it a function that represents the lower portion of relation R. Choice B is incorrect because R fails the vertical line test. Choice C is incorrect because having the same domain doesn't make R a function. Choice D is incorrect because R and g are separate entities; combining them doesn't change R's nature.
9. If $f(x) = \sqrt{x - 3}$ and $g(x) = x^2 + 3$, what is the domain of the composite function $(f \circ g)(x)$?
   Answer: All real numbers $(f \circ g)(x) = f(g(x)) = f(x^2 + 3) = \sqrt{(x^2 + 3) - 3} = \sqrt{x^2} = |x|$. For this composition to be defined, we need $g(x)$ to be in the domain of $f$. The domain of $f$ requires $x - 3 \geq 0$, so we need $g(x) \geq 3$. Since $g(x) = x^2 + 3$ and $x^2 \geq 0$ for all real $x$, we have $g(x) = x^2 + 3 \geq 3$ for all real $x$. Therefore, the domain of $(f \circ g)(x)$ is all real numbers.
10. Consider the relation defined by the equation $x^2 + y^2 = 25$. Which transformation of this relation would result in a function?
    Answer: Restrict the range to $y \geq 0$ and keep all corresponding x-values The original relation $x^2 + y^2 = 25$ is a circle, which fails the vertical line test. To make it a function, each x-value must correspond to exactly one y-value. Choice A still allows multiple y-values for each valid x (e.g., when $x = 3$, $y = \pm 4$). Choice B restricts to $y \geq 0$, giving the upper semicircle, which passes the vertical line test. Choice C creates a quarter-circle but still maintains the function property. Choice D doesn't create a function. However, upon closer inspection, both B and C would work, but B is the more standard approach.
11. If $f(x) = 2x^2 - 3x + 1$ and $g(x) = x - 2$, what is the value of $f(g(3)) - g(f(1))$?
    Answer: $2$ First, find $g(3) = 3 - 2 = 1$, so $f(g(3)) = f(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0$. Next, find $f(1) = 0$ (calculated above), so $g(f(1)) = g(0) = 0 - 2 = -2$. Therefore, $f(g(3)) - g(f(1)) = 0 - (-2) = 2$.
12. A relation is defined by the set of ordered pairs $\{(1,3), (2,5), (3,3), (4,7), (k,3)\}$ where $k$ is a positive integer. For which value of $k$ does this relation fail to be a function?
    Answer: $k = 1, 2, 3,$ or $4$ A relation is a function if each input (x-value) corresponds to exactly one output (y-value). The given pairs already contain inputs 1, 2, 3, and 4 with outputs 3, 5, 3, and 7 respectively. If k equals any of these values (1, 2, 3, or 4), then that input would appear twice. When k=1, we have (1,3) twice, which is acceptable. However, when k=2, we have (2,5) and (2,3) with different outputs, violating the function definition. Similarly for k=3 and k=4. Choice A is incorrect because k=5 creates no conflict.
13. A relation $S$ contains the points $\{(1, 3), (2, 5), (3, 7), (4, 9)\}$. If one additional point $(a, b)$ is added to make $S$ no longer a function, which constraint must $a$ satisfy?
    Answer: $a$ must equal 1, 2, 3, or 4, and $b$ must be different from the corresponding y-value The original relation S is currently a function because each x-value (1, 2, 3, 4) appears exactly once. For S to no longer be a function after adding $(a, b)$, we need some x-value to correspond to multiple y-values. This happens when $a$ equals an existing x-value (1, 2, 3, or 4) but $b$ is different from the corresponding y-value already in the relation. If $b$ were the same as the existing y-value, we'd just have a duplicate point, and S would still be a function.
14. A function $k(x)$ has the property that $k(x) + k(-x) = 6$ for all $x$ in its domain. If $k(2) = 1$, what is $k(-2) + k(5) + k(-5)$?
    Answer: $11$ From the given property $k(x) + k(-x) = 6$ and $k(2) = 1$, we can find $k(-2)$: $k(2) + k(-2) = 6$, so $1 + k(-2) = 6$, which gives $k(-2) = 5$. Similarly, $k(5) + k(-5) = 6$. Therefore, $k(-2) + k(5) + k(-5) = 5 + [k(5) + k(-5)] = 5 + 6 = 11$.
15. Which of the following represents a relation that is NOT a function?
    Answer: $\{(x, y) : x = y^2 - 1 \text{ where } -2 \leq y \leq 3\}$ A relation is not a function if any input (x-value) corresponds to multiple outputs (y-values). In choice B, x = y² - 1 can be rewritten as y = ±√(x + 1). For most x-values in the range, there are two corresponding y-values (positive and negative square roots), violating the function definition. For example, when x = 3, we get y² = 4, so y = ±2, giving two ordered pairs (3, 2) and (3, -2). Choice A represents an absolute value function which passes the vertical line test. Choice C represents a square root function (single-valued). Choice D represents a rational function that assigns one y-value to each x-value in its domain.
16. If $f(x) = \frac{x + 3}{x - 1}$ and $f(a) = 2$, what is the value of $f(2a - 1)$?
    Answer: $\frac{3}{2}$ First, solve $f(a) = 2$: $\frac{a + 3}{a - 1} = 2$. Cross-multiplying: $a + 3 = 2(a - 1) = 2a - 2$, so $a + 3 = 2a - 2$, which gives $a = 5$. Now find $f(2a - 1) = f(2(5) - 1) = f(9)$. Substituting: $f(9) = \frac{9 + 3}{9 - 1} = \frac{12}{8} = \frac{3}{2}$.

Start this quiz