Relations and Functions: Middle School Math Problems with Step-by-Step Solutions

The correct relation is “cube of”, as shown by the following relationships:
$1$ is the cube of $1$;
$8$ is the cube of $2$;
$27$ is the cube of $3$.
(Answer B)

The correct relation is “half of”, as shown by the following relationships:
$-3$ is half of $-6$;
$-1$ is half of $-2$;
$1$ is half of $2$;
$2$ is half of $4$;
(Answer B)

The correct relation is “half of”, as shown by the following relationships:
$1$ is half of $2$;
$2$ is half of $4$;
$3$ is half of $6$;
$4$ is half of $8$, but since $8$ is not a member of the codomain, $4$ does not have a pair.
(Answer C)

The arrow diagram above can be expressed in the form of ordered pairs $\{1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (4, 4)\}$.
Notice that $1$ is a factor of $2, 3$, and $4$; $2$ is a factor of $2$ and $4$; $4$ is a factor of $4$.
Therefore, the appropriate relation is FACTOR OF.
Note: a factor is a natural number that divides another number exactly.
(Answer A)

In the ordered pair form $(a, b)$, $a$ is called a domain member, while $b$ is called a range member.
A relation is called a mapping (function) if every domain member has exactly one (must be one) pair in the codomain.
In set (1), all domain members have exactly one pair in the codomain, so it is called a function.
In set (2), the domain member $2$ has two pairs, so it is not a function.
In set (3), the domain member $3$ has two pairs, so it is not a function.
In set (4), the domain member $3$ has two pairs, so it is not a function.
(Answer A)

A set of ordered pairs is considered a function if every domain member has exactly one pair in the codomain.
Set $P$ is not a function because the domain member $2$ has more than one pair, namely in the ordered pairs $(2, 0)$ and $(2, 1)$.
Set $Q$ is a function because every domain member has exactly one pair in the codomain.
Set $R$ is not a function because the domain member $5$ has more than one pair, namely in the ordered pairs $(5, a)$ and $(5, b)$.
Set $S$ is not a function because the domain member $1$ has more than one pair, namely in the ordered pairs $(1, 6), (1, 5)$ and $(1, 4)$.
(Answer B)

Given:
$\begin{aligned} A & = \{a, b, c\} \\ B & = \{1,2,3,4,5\} \end{aligned}$
We obtain $\text{n}(A) = 3$ and $\text{n}(B) = 5$, therefore the number of possible mappings from $A$ to $B$ is $\boxed{\text{n}(B)^{\text{n}(A)} = 5^3 = 125}$
(Answer C)

First, write down all members of sets $A$ and $B$.
$\begin{aligned} A & = \{1, 2, 3, 6\} \\ B & = \{2, 4, 6\} \end{aligned}$
We obtain $\text{n}(A) = 4$ and $\text{n}(B) = 3$, therefore the number of possible mappings from $B$ to $A$ is $\boxed{\text{n}(A)^{\text{n}(B)} = 4^3 = 64}$
(Answer B)

Given: $f(x) = 3-5x$.
Substitute $x =-4$ to obtain
$f(-4) = 3-5(-4) = 3 + 20 = 23$
Therefore, the result is $\boxed{f(-4) = 23}$
(Answer D)

Given: $f(x) = 6-3x$.
Substitute $x = 5$ to obtain
$f(5) = 6-3(5) = 6-15 =-9$
Substitute $x =-4$ to obtain
$f(-4) = 6-3(-4) = 6 + 12 = 18$
Therefore, the result is $\boxed{f(5) + f(-4) =-9 + 18 = 9}$
(Answer B)

Substitute the value $x =-\dfrac{1}{2}$ into the function $f$.
$\begin{aligned} f(x) & = 3x^2-2x-5 \\ f\left(-\dfrac{1}{2}\right) & = 3\left(-\dfrac{1}{2}\right)^2-2\left(-\dfrac{1}{2}\right)-5 \\ & = 3 \times \dfrac{1}{4} + 1-5 \\ & = \dfrac{3}{4}-4 \\ & =-3\dfrac{1}{4} \end{aligned}$
Therefore, the value of $\boxed{f\left(-\dfrac{1}{2}\right) =-3\dfrac{1}{4}}$
(Answer B)

Given: $f(x) = 6x- 5$.
Substitute $x = 3a + 1$ to obtain
$\begin{aligned} f(3a+1) & = 6(3a+1)-5 \\ & = 18a+6-5 \\ & = 18a+1 \end{aligned}$
Therefore, the result is $\boxed{f(3a+1) = 18a+1}$
(Answer C)

Given: $f(x) = 3x + 5$.
Substitute $x = 2b-3$ to obtain
$\begin{aligned} f(2b-3) & = 3(2b-3) + 5 \\ & = 6b-9 + 5 \\ & = 6b-4 \end{aligned}$
Therefore, the result is $\boxed{f(2b-3) = 6b- 4}$
(Answer C)

Given: $f(3x+2)=6x+10$
We will determine the value of $x$ that makes $3x + 2 =-5$ as follows.
$\begin{aligned} 3x + 2 & =-5 \\ 3x & =-7 \\ x & =-\dfrac{7}{3} \end{aligned}$
Substitute $x =-\dfrac73$ into $f(x)$ to obtain
$\begin{aligned} f\left(\cancel{3}\left(-\dfrac{7}{\cancel{3}}\right) + 2\right) & = \cancelto{2}{6}\left(-\dfrac{7}{\cancel{3}}\right) + 10 \\ f(-7+2) & = 2(-7) + 10 \\ f(-5) & =-4 \end{aligned}$
Therefore, the value of $\boxed{f(-5) =-4}$
(Answer B)

Given: $f(3x+3)=(x-4)(x+10)$
We will determine the value of $x$ that makes $3x + 3 = 21$ as follows.
$\begin{aligned} 3x + 3 & = 21 \\ 3x & = 18 \\ x & = \dfrac{18}{3} = 6 \end{aligned}$
Substitute $x = 6$ into $f(x)$ to obtain
$\begin{aligned} f(3(6)+3) & = (6-4)(6+10) \\ f(18+3) & = 2 \times 16 \\ f(21) & = 32 \end{aligned}$
Therefore, the value of $\boxed{f(21) = 32}$
(Answer C)

Given: $f(x) = 2x + 5$ and $f(a) = 11$
Substitute $x = a$ to obtain
$\begin{aligned} f(a) & = 2a + 5 \\ 11 & = 2a + 5 \\ 6 & = 2a \\ a & = 3 \end{aligned}$
Therefore, the value of $a$ is $\boxed{3}$
(Answer B)

Given: $f(x) = 4x- 1$.
Substitute $x = a$ to obtain
$\begin{aligned} f(a) & = 4a-1 \\-9 & = 4a-1 \\-8 & = 4a \\ a & =-2 \end{aligned}$
Substitute $x = 2$ to obtain
$f(2) = 4(2)-1 = 8-1 = 7$
This means that the value of $b$ is $7$.
Therefore, the result is $\boxed{a-b =-2-7 =-9}.$
(Answer A)

Given: $f(x) = 4x + 3$.
Substitute $x = a$ to obtain
$\begin{aligned} f(a) & = 4a + 3 \\ 7 & = 4a + 3 \\ 4 & = 4a \\ a & = 1 \end{aligned}$
Substitute $x =-2$ to obtain
$f(2) = 4(-2) + 3 =-8+3 =-5$
This means that the value of $b$ is $-5$.
Therefore, the result is $\boxed{a+b = 1 +(-5) =-4}$
(Answer C)

Given: $f(x)= 2x-3$.
For $x = m$, we obtain
$\begin{aligned} f(m) & = 2m-3 \\ 5 & = 2m-3 \\ 8 & = 2m \\ m & = 4 \end{aligned}$
For $x =-2$, we obtain
$f(-2) = 2(-2)-3 =-7$
thus the value of $n$ is $-7$.
Therefore, the value of $\boxed{m+n=4+(-7)=-3}$
(Answer C)

For $x = 2$, we obtain
$f(2) = 2p + q = 5.$
For $x =-2$, we obtain
$f(-2) =-2p + q =-11.$
We obtain the system of linear equations: $\begin{cases} 2p+q = 5 \\-2p+q=-11 \end{cases}$
Using the elimination-substitution method, we obtain
$\begin{aligned} \! \begin{aligned} 2p + q & = 5 \\-2p+q & =-11 \end{aligned} \\ \rule{2.5 cm}{0.8pt}- \\ \! \begin{aligned} 4p & = 16 \\ p & = 4 \end{aligned} \end{aligned}$
Next, substitute $p=4$ into one of the equations, for example $2p+q=5.$
$2(4)+q = 5 \Leftrightarrow q =-3.$
Therefore, the values are $\boxed{p=4}$ and $\boxed{q=-3}$.
Thus, $f(x) = 4x-3.$
For $x=-6$, we obtain
$\boxed{f(-6) = 4(-6)-3 =-27}$
(Answer A)

Test each provided option.
Option A: $f(x) = 2x + 7$
For $x = 6$, we obtain $f(6) = 2(6) + 7 = 19$ (incorrect), it should be $15$.
Option B: $f(x) = 5x- 12$
For $x = 6$, we obtain $f(6) = 5(6)-12 = 18$ (incorrect), it should be $15$.
Option C: $f(x) = 3x- 2$
For $x = 6$, we obtain $f(6) = 3(6)-2= 16$ (incorrect), it should be $15$.
Option D: $f(x) = 2x + 3$
For $x = 5$, we obtain $f(5) = 2(5) + 3 = 13$ (correct).
For $x = 6$, we obtain $f(5) = 2(6) + 3 = 15$ (correct).
For $x = 8$, we obtain $f(5) = 2(8) + 3 = 19$ (correct).
(Answer D)

Test each provided option.
Option A: $f(x) = 4(2x + 5)$
For $x = 2$, we obtain $f(2) = 4(2(2) + 5) = 4(9) = 36$ (incorrect), it should be $21$.
Option B: $f(x) = 3(2x+3)$
For $x = 2$, we obtain $f(2) = 3(2(2) + 3) = 3(7) = 21$ (correct).
For $x = 6$, we obtain $f(6) = 3(2(6) + 3) = 3(15) = 45$ (correct).
For $x = 10$, we obtain $f(10) = 3(2(10) + 3) = 3(23) = 69$ (correct).
Option C: $f(x) = 2(3x + 9)$
For $x = 2$, we obtain $f(2) = 2(3(2) + 9) = 2(15) = 30$ (incorrect), it should be $21$.
Option D: $f(x) =\dfrac12(6x + 18) = 3x + 9$
For $x = 2$, we obtain $f(2) = 3(2) + 9 = 15$ (incorrect), it should be $21$.
(Answer B)

Given $f(x) = \dfrac{1}{a}x+b$ and $h(x)=bx-a$. Since the graphs of $f(x)$ and $h(x)$ are parallel, their gradients must be equal, namely $m_f = m_g \Rightarrow \dfrac{1}{a} = b$, equivalent to $ab = 1$.
Next, we obtain
$\begin{aligned} \dfrac{h(-1)}{f(1)} & = -\dfrac52 \\ \dfrac{b(-1)-a}{\dfrac{1}{a}(1)+b} & = -\dfrac52 \\ \dfrac{-a-b}{\dfrac{1}{a}+b} & = -\dfrac52 \\ \text{Multiply both}~&\text{sides by}~-1 \\ \dfrac{a+b}{\dfrac{1}{a}+b} \color{blue}{\times \dfrac{a}{a}} & = \dfrac52 \\ \dfrac{a^2+ab}{1+ab} & = \dfrac52 \\ \text{Substitute}~& ab = 1 \\ \dfrac{a^2+1}{1+1} & = \dfrac52 \\ \dfrac{a^2+1}{2} & = \dfrac52 \\ a^2+1 & = 5 \\ a^2 & = 4 \\ a & = \pm 2 \end{aligned}$
We obtain two possible values for $a$, namely $a = 2$ or $a = -2$. From $ab = 1$, we can determine the value of $b$.

1. If $a = 2$, then we obtain $b = \dfrac12$ so that $b-a = \dfrac12-2 = -\dfrac32$.
2. If $a = -2$, then we obtain $b = -\dfrac12$ so that $b-a=-\dfrac12-(-2) = \dfrac32$.

Thus, there will be two values for $b-a$, namely $\boxed{\pm \dfrac32}$
(Answer B)

Given $H(\sqrt{x+5}) = x$.
Change $x \to x-5$ so that we obtain
$\begin{aligned} H\left(\sqrt{(x-5)+5}\right) & = x-5 \\ H(\sqrt{x}) = x-5 \end{aligned}$
Change $x \to x^4$ so that we obtain
$\begin{aligned} H(\sqrt{x^4}) & = x^4-5 \\ H(x^2) & = x^4-5 \end{aligned}$
Thus, we obtain that $a=4$ and $b=5$ so that the value of $\boxed{a+b=4+5=9}$
(Answer B)

A curve drawn on the Cartesian plane is said to be a function if each value of $x$ corresponds to exactly one value of $y$. Note that the horizontal axis is called the $X$-axis, while the vertical axis is called the $Y$-axis.
Simply put, we draw a vertical dashed line intersecting the curve. If we find more than one intersection point or no intersection point at all, then the curve is not the graph of a function.

Answer a)
We only find one intersection point when drawing a dashed line from any position. Therefore, the curve represents a function.

Answer b)
We only find one intersection point when drawing a dashed line from any position. Therefore, the curve represents a function.

Answer c)
We find two intersection points when drawing a dashed line from the position shown in the figure. Therefore, the curve does not represent a function.

Answer d)
We find two intersection points when drawing a dashed line from the position shown in the figure. Therefore, the curve does not represent a function.

Answer e)
We find two intersection points when drawing a dashed line from the position shown in the figure. Therefore, the curve does not represent a function.
Answer f)

We find three intersection points when drawing a dashed line from the position shown in the figure. Therefore, the curve does not represent a function.
Answer g)
We find two intersection points when drawing a dashed line from the position shown in the figure. In addition, if we draw a dashed line coinciding with the $Y$-axis, we will not find any intersection point. Therefore, the curve does not represent a function.