Functions and Relations: Explanation and Examples

Introduction to Functions and Relations

In mathematics, relations and functions are fundamental concepts used to describe how elements from one set relate to elements of another set.

A relation is simply a set of ordered pairs, where the first element comes from one set (called the domain), and the second element comes from another set (called the range). Relations can describe any possible pairing between elements of two sets.

A function is a special kind of relation where each element from the domain (the input) is related to exactly one element in the range (the output). In simpler terms, a function assigns exactly one output to each input.

Definitions

• Domain: The set of all possible input values.
• Range: The set of all possible output values.
• Relation: A set of ordered pairs (x, y), where $x$ is from the domain and $y$ is from the range.
• Function: A relation in which each element of the domain is paired with exactly one element in the range.

Function Notation

If $f$ is a function and $x$ is an element from the domain, then $f(x)$ represents the output or value of the function at $x$. The equation $y=f(x)$ means that the value of $y$ depends on the value of $x$ according to the rule defined by $f$.

---

Example 1:

Let’s consider the relation $R=(1,2),(2,4),(3,6),(4,8)$. Is this a function?

Solution:

Step 1: Check if each element in the domain has exactly one corresponding element in the range.

For $x=1$, $y=2$
For $x=2$, $y=4$
For $x=3$, $y=6$
For $x=4$, $y=8$

Since each input $x$ corresponds to only one output $y$, this relation is a function.

---

Example 2:

Consider the relation $R=(1,2),(2,3),(2,4)$. Is this a function?

Solution:

Step 1: Check if each element in the domain has exactly one corresponding element in the range.

For $x=1$, $y=2$
For $x=2$, there are two outputs, $y=3$ and $y=4$.

Since $x=2$ corresponds to two different values of $y$, this relation is not a function.

---

Example 3:

Let $f(x)=x^2+2x+1$. Find $f(3)$.

Solution:

Step 1: Substitute $x=3$ into the function.

$f(3)=3^2+2(3)+1$
$=9+6+1$
$=16$

Thus, $f(3)=16$.

---

Example 4:

Let $f(x)=2x–5$. Find $f(-1)$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=2(-1)–5$
$=-2–5$
$=-7$

Thus, $f(-1)=-7$.

---

Example 5:

Let $f(x)=x^3–x$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=0^3–0$
$=0$

Thus, $f(0)=0$.

---

Example 6:

Consider the relation $R=(1,2),(2,3),(3,4)$. Find the domain and range of this relation.

Solution:

Step 1: The domain is the set of all first elements in the ordered pairs: $1,2,3$.

Step 2: The range is the set of all second elements in the ordered pairs: $2,3,4$.

Thus, the domain is $1,2,3$ and the range is $2,3,4$.

---

Example 7:

Determine if the following is a function: $f=(1,2),(2,4),(3,6),(4,8),(2,5)$.

Solution:

Step 1: Check if any $x$ value corresponds to more than one $y$ value.

For $x=2$, there are two outputs: $y=4$ and $y=5$.

Since there is more than one output for $x=2$, this is not a function.

---

Example 8:

Find the range of the function $f(x)=x^2$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=1^2=1$
$f(2)=2^2=4$
$f(3)=3^2=9$
$f(4)=4^2=16$

Thus, the range is $1,4,9,16$.

---

Example 9:

Is the relation $R=(0,1),(1,0),(2,-1),(0,1)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

The $x$ value $0$ appears twice with the same $y$ value $1$. Since no $x$ value has more than one output, this is a function.

---

Example 10:

Let $g(x)=x^2–4x+3$. Find $g(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$g(2)=2^2–4(2)+3$
$=4–8+3$
$=-1$

Thus, $g(2)=-1$.

---

Example 11:

Let $f(x)=3x+2$. Find $f(4)$.

Solution:

Step 1: Substitute $x=4$ into the function.

$f(4)=3(4)+2$
$=12+2$
$=14$

Thus, $f(4)=14$.

---

Example 12:

Find the domain and range of the function $f(x)=2x+1$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0)+1=1$
$f(1)=2(1)+1=3$
$f(2)=2(2)+1=5$
$f(3)=2(3)+1=7$

Thus, the domain is $0,1,2,3$ and the range is $1,3,5,7$.

---

Example 13:

Is the relation $R=(1,2),(2,3),(3,4),(1,5)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs, $y=2$ and $y=5$. Since one $x$ has more than one $y$, this is not a function.

---

Example 14:

Find $f(x)=x^2–4x+6$ for $x=0$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=0^2–4(0)+6$
$=0–0+6$
$=6$

Thus, $f(0)=6$.

---

Example 15:

Determine whether $R=(1,4),(2,8),(3,12)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For each $x$ value (1, 2, and 3), there is only one $y$ value. Therefore, this is a function.

---

Example 16:

Find the range of the function $f(x)=2x–3$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=2(1)–3=-1$
$f(2)=2(2)–3=1$
$f(3)=2(3)–3=3$
$f(4)=2(4)–3=5$

Thus, the range is $-1,1,3,5$.

---

Example 17:

Let $g(x)=x^3–x$. Find $g(-1)$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$g(-1)=(-1{)}^3–(-1)$
$=-1+1$
$=0$

Thus, $g(-1)=0$.

---

Example 18:

Is the relation $R=(0,1),(2,3),(3,5),(3,6)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=3$, there are two outputs, $y=5$ and $y=6$. Since one $x$ has more than one $y$, this is not a function.

---

Example 19:

Find the domain and range of the function $f(x)=x^2+2x$ for the domain $-2,-1,0,1$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-2)=(-2{)}^2+2(-2)=4–4=0$
$f(-1)=(-1{)}^2+2(-1)=1–2=-1$
$f(0)=0^2+2(0)=0$
$f(1)=1^2+2(1)=1+2=3$

Thus, the domain is $-2,-1,0,1$ and the range is $0,-1,3$.

---

Example 20:

Let $f(x)=4x–7$. Find $f(5)$.

Solution:

Step 1: Substitute $x=5$ into the function.

$f(5)=4(5)–7$
$=20–7$
$=13$

Thus, $f(5)=13$.

---

Example 21:

Find the value of $f(x)=3x^2–2x+1$ for $x=2$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=3(2{)}^2–2(2)+1$
$=3(4)–4+1$
$=12–4+1$
$=9$

Thus, $f(2)=9$.

---

Example 22:

Let $f(x)=x^2+4x+3$. Find the value of $f(-1)$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=(-1{)}^2+4(-1)+3$
$=1–4+3$
$=0$

Thus, $f(-1)=0$.

---

Example 23:

Determine if the relation $R=(2,5),(3,6),(4,7),(3,8)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=3$, there are two outputs, $y=6$ and $y=8$. Since one $x$ has more than one $y$, this is not a function.

---

Example 24:

Find the range of the function $f(x)=2x–3$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0)–3=-3$
$f(1)=2(1)–3=-1$
$f(2)=2(2)–3=1$
$f(3)=2(3)–3=3$

Thus, the range is $-3,-1,1,3$.

---

Example 25:

Let $f(x)=x^3–3x^2+2$. Find $f(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$f(1)=(1{)}^3–3(1{)}^2+2$
$=1–3+2$
$=0$

Thus, $f(1)=0$.

---

Example 26:

Is the relation $R=(1,4),(2,5),(1,6)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs, $y=4$ and $y=6$. Since one $x$ has more than one $y$, this is not a function.

---

Example 27:

Find the domain and range of the function $f(x)=3x–2$ for the domain $2,3,4,5$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(2)=3(2)–2=6–2=4$
$f(3)=3(3)–2=9–2=7$
$f(4)=3(4)–2=12–2=10$
$f(5)=3(5)–2=15–2=13$

Thus, the domain is $2,3,4,5$ and the range is $4,7,10,13$.

---

Example 28:

Let $g(x)=4x^2–5x+6$. Find $g(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$g(0)=4(0{)}^2–5(0)+6$
$=0+0+6$
$=6$

Thus, $g(0)=6$.

---

Example 29:

Is the relation $R=(2,3),(4,5),(6,7)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

Each $x$ value corresponds to one $y$ value. Therefore, this is a function.

---

Example 30:

Find the range of the function $f(x)=x^2–2x+1$ for the domain $-1,0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-1)=(-1{)}^2–2(-1)+1=1+2+1=4$
$f(0)=(0{)}^2–2(0)+1=0+0+1=1$
$f(1)=(1{)}^2–2(1)+1=1–2+1=0$
$f(2)=(2{)}^2–2(2)+1=4–4+1=1$

Thus, the range is $4,1,0$.

---

Example 31:

Find $f(x)=x^2+5x–3$ for $x=4$.

Solution:

Step 1: Substitute $x=4$ into the function.

$f(4)=(4{)}^2+5(4)–3$
$=16+20–3$
$=33$

Thus, $f(4)=33$.

---

Example 32:

Let $f(x)=2x^2–x+4$. Find $f(-2)$.

Solution:

Step 1: Substitute $x=-2$ into the function.

$f(-2)=2(-2{)}^2–(-2)+4$
$=2(4)+2+4$
$=8+2+4$
$=14$

Thus, $f(-2)=14$.

---

Example 33:

Let $f(x)=3x^2+4x–5$. Find $f(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$f(1)=3(1{)}^2+4(1)–5$
$=3+4–5$
$=2$

Thus, $f(1)=2$.

---

Example 34:

Find the domain and range of the function $f(x)=2x+3$ for the domain $-1,0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-1)=2(-1)+3=-2+3=1$
$f(0)=2(0)+3=0+3=3$
$f(1)=2(1)+3=2+3=5$
$f(2)=2(2)+3=4+3=7$

Thus, the domain is $-1,0,1,2$ and the range is $1,3,5,7$.

---

Example 35:

Is the relation $R=(2,4),(2,6),(3,5)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs: $y=4$ and $y=6$. Since one $x$ has more than one $y$, this is not a function.

---

Example 36:

Find the value of $f(x)=x^3–2x+1$ for $x=0$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=(0{)}^3–2(0)+1$
$=0–0+1$
$=1$

Thus, $f(0)=1$.

---

Example 37:

Find the range of the function $f(x)=x^2+3x–1$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=(1{)}^2+3(1)–1=1+3–1=3$
$f(2)=(2{)}^2+3(2)–1=4+6–1=9$
$f(3)=(3{)}^2+3(3)–1=9+9–1=17$
$f(4)=(4{)}^2+3(4)–1=16+12–1=27$

Thus, the range is $3,9,17,27$.

---

Example 38:

Let $f(x)=x^2+2x$. Find $f(3)$.

Solution:

Step 1: Substitute $x=3$ into the function.

$f(3)=(3{)}^2+2(3)$
$=9+6$
$=15$

Thus, $f(3)=15$.

---

Example 39:

Determine if the relation $R=(0,1),(1,3),(2,1),(2,4)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs, $y=1$ and $y=4$. Since one $x$ has more than one $y$, this is not a function.

---

Example 40:

Find the domain and range of the function $f(x)=3x–5$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=3(0)–5=-5$
$f(1)=3(1)–5=-2$
$f(2)=3(2)–5=1$
$f(3)=3(3)–5=4$

Thus, the domain is $0,1,2,3$ and the range is $-5,-2,1,4$.

---

Example 41:

Let $f(x)=x^2–4x+4$. Find $f(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=(2{)}^2–4(2)+4$
$=4–8+4$
$=0$

Thus, $f(2)=0$.

---

Example 42:

Find the value of $f(x)=5x–3$ for $x=-2$.

Solution:

Step 1: Substitute $x=-2$ into the function.

$f(-2)=5(-2)–3$
$=-10–3$
$=-13$

Thus, $f(-2)=-13$.

---

Example 43:

Find the range of the function $f(x)=x^2+3x$ for the domain $1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=(1{)}^2+3(1)=1+3=4$
$f(2)=(2{)}^2+3(2)=4+6=10$
$f(3)=(3{)}^2+3(3)=9+9=18$

Thus, the range is $4,10,18$.

---

Example 44:

Is the relation $R=(1,2),(2,3),(3,4),(2,5)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs: $y=3$ and $y=5$. Since one $x$ has more than one $y$, this is not a function.

---

Example 45:

Find the domain and range of the function $f(x)=4x–7$ for the domain $2,3,4,5$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(2)=4(2)–7=8–7=1$
$f(3)=4(3)–7=12–7=5$
$f(4)=4(4)–7=16–7=9$
$f(5)=4(5)–7=20–7=13$

Thus, the domain is $2,3,4,5$ and the range is $1,5,9,13$.

---

Example 46:

Let $g(x)=x^3–2x+5$. Find $g(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$g(1)=(1{)}^3–2(1)+5$
$=1–2+5$
$=4$

Thus, $g(1)=4$.

---

Example 47:

Find the value of $f(x)=2x^2–3x+1$ for $x=0$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=2(0{)}^2–3(0)+1$
$=0+0+1$
$=1$

Thus, $f(0)=1$.

---

Example 48:

Let $f(x)=3x^2+2x–7$. Find $f(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=3(2{)}^2+2(2)–7$
$=3(4)+4–7$
$=12+4–7$
$=9$

Thus, $f(2)=9$.

---

Example 49:

Find the domain and range of the function $f(x)=4x+2$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=4(0)+2=2$
$f(1)=4(1)+2=6$
$f(2)=4(2)+2=10$
$f(3)=4(3)+2=14$

Thus, the domain is $0,1,2,3$ and the range is $2,6,10,14$.

---

Example 50:

Determine if the relation $R=(1,2),(2,3),(3,4),(1,5)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs, $y=2$ and $y=5$. Since one $x$ has more than one $y$, this is not a function.

---

Example 51:

Find the value of $f(x)=x^2–5x+6$ for $x=-1$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=(-1{)}^2–5(-1)+6$
$=1+5+6$
$=12$

Thus, $f(-1)=12$.

---

Example 52:

Find the range of the function $f(x)=2x^2–3x$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0{)}^2–3(0)=0$
$f(1)=2(1{)}^2–3(1)=2–3=-1$
$f(2)=2(2{)}^2–3(2)=8–6=2$
$f(3)=2(3{)}^2–3(3)=18–9=9$

Thus, the range is $0,-1,2,9$.

---

Example 53:

Let $f(x)=x^2+4x+4$. Find $f(-2)$.

Solution:

Step 1: Substitute $x=-2$ into the function.

$f(-2)=(-2{)}^2+4(-2)+4$
$=4–8+4$
$=0$

Thus, $f(-2)=0$.

---

Example 54:

Find the domain and range of the function $f(x)=5x–2$ for the domain $-1,0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-1)=5(-1)–2=-5–2=-7$
$f(0)=5(0)–2=0–2=-2$
$f(1)=5(1)–2=5–2=3$
$f(2)=5(2)–2=10–2=8$

Thus, the domain is $-1,0,1,2$ and the range is $-7,-2,3,8$.

---

Example 55:

Is the relation $R=(2,3),(3,5),(4,7),(2,6)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs: $y=3$ and $y=6$. Since one $x$ has more than one $y$, this is not a function.

---

Example 56:

Let $f(x)=x^3–4x+1$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=(0{)}^3–4(0)+1$
$=0+0+1$
$=1$

Thus, $f(0)=1$.

---

Example 57:

Find the range of the function $f(x)=3x+2$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=3(1)+2=3+2=5$
$f(2)=3(2)+2=6+2=8$
$f(3)=3(3)+2=9+2=11$
$f(4)=3(4)+2=12+2=14$

Thus, the range is $5,8,11,14$.

---

Example 58:

Let $f(x)=x^2–x+3$. Find $f(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=(2{)}^2–2(2)+3$
$=4–4+3$
$=3$

Thus, $f(2)=3$.

---

Example 59:

Find the domain and range of the function $f(x)=2x–4$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0)–4=-4$
$f(1)=2(1)–4=-2$
$f(2)=2(2)–4=0$
$f(3)=2(3)–4=2$

Thus, the domain is $0,1,2,3$ and the range is $-4,-2,0,2$.

---

Example 60:

Let $g(x)=x^2+x–1$. Find $g(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$g(1)=(1{)}^2+1–1$
$=1+1–1$
$=1$

Thus, $g(1)=1$.

---

Example 61:

Find the value of $f(x)=4x–3$ for $x=-1$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=4(-1)–3$
$=-4–3$
$=-7$

Thus, $f(-1)=-7$.

---

Example 62:

Find the domain and range of the function $f(x)=2x+1$ for the domain $-1,0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-1)=2(-1)+1=-2+1=-1$
$f(0)=2(0)+1=0+1=1$
$f(1)=2(1)+1=2+1=3$
$f(2)=2(2)+1=4+1=5$

Thus, the domain is $-1,0,1,2$ and the range is $-1,1,3,5$.

---

Example 63:

Let $f(x)=3x^2–2x+5$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=3(0{)}^2–2(0)+5$
$=0–0+5$
$=5$

Thus, $f(0)=5$.

---

Example 64:

Find the range of the function $f(x)=x^2–3x+4$ for the domain $1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=(1{)}^2–3(1)+4=1–3+4=2$
$f(2)=(2{)}^2–3(2)+4=4–6+4=2$
$f(3)=(3{)}^2–3(3)+4=9–9+4=4$

Thus, the range is $2,4$.

---

Example 65:

Is the relation $R=(0,1),(2,3),(3,4),(0,2)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=0$, there are two outputs: $y=1$ and $y=2$. Since one $x$ has more than one $y$, this is not a function.

---

Example 66:

Let $f(x)=x^3–4x$. Find $f(-1)$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=(-1{)}^3–4(-1)$
$=-1+4$
$=3$

Thus, $f(-1)=3$.

---

Example 67:

Find the domain and range of the function $f(x)=2x–7$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0)–7=-7$
$f(1)=2(1)–7=-5$
$f(2)=2(2)–7=-3$
$f(3)=2(3)–7=-1$

Thus, the domain is $0,1,2,3$ and the range is $-7,-5,-3,-1$.

---

Example 68:

Let $f(x)=4x^2+3x–2$. Find $f(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$f(1)=4(1{)}^2+3(1)–2$
$=4+3–2$
$=5$

Thus, $f(1)=5$.

---

Example 69:

Find the range of the function $f(x)=2x^2–3x+1$ for the domain $1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=2(1{)}^2–3(1)+1=2–3+1=0$
$f(2)=2(2{)}^2–3(2)+1=8–6+1=3$
$f(3)=2(3{)}^2–3(3)+1=18–9+1=10$

Thus, the range is $0,3,10$.

---

Example 70:

Is the relation $R=(1,2),(2,4),(2,6)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs: $y=4$ and $y=6$. Since one $x$ has more than one $y$, this is not a function.

---

Example 71:

Let $f(x)=x^2+5x+6$. Find $f(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=(2{)}^2+5(2)+6$
$=4+10+6$
$=20$

Thus, $f(2)=20$.

---

Example 72:

Find the value of $f(x)=4x^2–5x+2$ for $x=0$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=4(0{)}^2–5(0)+2$
$=0+0+2$
$=2$

Thus, $f(0)=2$.

---

Example 73:

Find the range of the function $f(x)=3x–1$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=3(1)–1=3–1=2$
$f(2)=3(2)–1=6–1=5$
$f(3)=3(3)–1=9–1=8$
$f(4)=3(4)–1=12–1=11$

Thus, the range is $2,5,8,11$.

---

Example 74:

Let $f(x)=x^2–6x+8$. Find $f(3)$.

Solution:

Step 1: Substitute $x=3$ into the function.

$f(3)=(3{)}^2–6(3)+8$
$=9–18+8$
$=-1$

Thus, $f(3)=-1$.

---

Example 75:

Find the domain and range of the function $f(x)=5x–6$ for the domain $-2,-1,0,1$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-2)=5(-2)–6=-10–6=-16$
$f(-1)=5(-1)–6=-5–6=-11$
$f(0)=5(0)–6=0–6=-6$
$f(1)=5(1)–6=5–6=-1$

Thus, the domain is $-2,-1,0,1$ and the range is $-16,-11,-6,-1$.

---

Example 76:

Let $f(x)=x^3–x+2$. Find $f(-2)$.

Solution:

Step 1: Substitute $x=-2$ into the function.

$f(-2)=(-2{)}^3–(-2)+2$
$=-8+2+2$
$=-4$

Thus, $f(-2)=-4$.

---

Example 77:

Find the range of the function $f(x)=2x+5$ for the domain $0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=2(0)+5=5$
$f(1)=2(1)+5=7$
$f(2)=2(2)+5=9$

Thus, the range is $5,7,9$.

---

Example 78:

Is the relation $R=(1,2),(2,4),(1,3)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs: $y=2$ and $y=3$. Since one $x$ has more than one $y$, this is not a function.

---

Example 79:

Let $f(x)=2x^2–4x+1$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=2(0{)}^2–4(0)+1$
$=0+0+1$
$=1$

Thus, $f(0)=1$.

---

Example 80:

Find the domain and range of the function $f(x)=x^2+2x+1$ for the domain $-2,-1,0,1$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(-2)=(-2{)}^2+2(-2)+1=4–4+1=1$
$f(-1)=(-1{)}^2+2(-1)+1=1–2+1=0$
$f(0)=(0{)}^2+2(0)+1=0+0+1=1$
$f(1)=(1{)}^2+2(1)+1=1+2+1=4$

Thus, the domain is $-2,-1,0,1$ and the range is $0,1,4$.

---

Example 81:

Let $f(x)=x^2–2x+3$. Find $f(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$f(1)=(1{)}^2–2(1)+3$
$=1–2+3$
$=2$

Thus, $f(1)=2$.

---

Example 82:

Find the range of the function $f(x)=3x^2–4x+1$ for the domain $0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=3(0{)}^2–4(0)+1=1$
$f(1)=3(1{)}^2–4(1)+1=3–4+1=0$
$f(2)=3(2{)}^2–4(2)+1=12–8+1=5$

Thus, the range is $1,0,5$.

---

Example 83:

Is the relation $R=(1,2),(2,3),(3,2),(1,4)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs: $y=2$ and $y=4$. Since one $x$ has more than one $y$, this is not a function.

---

Example 84:

Let $f(x)=2x^2+3x–5$. Find $f(-1)$.

Solution:

Step 1: Substitute $x=-1$ into the function.

$f(-1)=2(-1{)}^2+3(-1)–5$
$=2(1)–3–5$
$=2–3–5$
$=-6$

Thus, $f(-1)=-6$.

---

Example 85:

Find the domain and range of the function $f(x)=4x–3$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=4(1)–3=4–3=1$
$f(2)=4(2)–3=8–3=5$
$f(3)=4(3)–3=12–3=9$
$f(4)=4(4)–3=16–3=13$

Thus, the domain is $1,2,3,4$ and the range is $1,5,9,13$.

---

Example 86:

Let $f(x)=x^3–2x+1$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=(0{)}^3–2(0)+1$
$=0+0+1$
$=1$

Thus, $f(0)=1$.

---

Example 87:

Find the range of the function $f(x)=2x–4$ for the domain $1,2,3,4$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=2(1)–4=2–4=-2$
$f(2)=2(2)–4=4–4=0$
$f(3)=2(3)–4=6–4=2$
$f(4)=2(4)–4=8–4=4$

Thus, the range is $-2,0,2,4$.

---

Example 88:

Determine if the relation $R=(1,2),(2,3),(3,4),(3,5)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=3$, there are two outputs: $y=4$ and $y=5$. Since one $x$ has more than one $y$, this is not a function.

---

Example 89:

Let $f(x)=x^2+5x+6$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=(0{)}^2+5(0)+6$
$=0+0+6$
$=6$

Thus, $f(0)=6$.

---

Example 90:

Find the range of the function $f(x)=3x^2–2x+1$ for the domain $0,1,2$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=3(0{)}^2–2(0)+1=1$
$f(1)=3(1{)}^2–2(1)+1=3–2+1=2$
$f(2)=3(2{)}^2–2(2)+1=12–4+1=9$

Thus, the range is $1,2,9$.

---

Example 91:

Let $f(x)=x^3–3x^2+2$. Find $f(1)$.

Solution:

Step 1: Substitute $x=1$ into the function.

$f(1)=(1{)}^3–3(1{)}^2+2$
$=1–3+2$
$=0$

Thus, $f(1)=0$.

---

Example 92:

Find the domain and range of the function $f(x)=5x–7$ for the domain $1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=5(1)–7=5–7=-2$
$f(2)=5(2)–7=10–7=3$
$f(3)=5(3)–7=15–7=8$

Thus, the domain is $1,2,3$ and the range is $-2,3,8$.

---

Example 93:

Is the relation $R=(0,1),(2,3),(2,4)$ a function?

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=2$, there are two outputs: $y=3$ and $y=4$. Since one $x$ has more than one $y$, this is not a function.

---

Example 94:

Let $f(x)=x^2–x+4$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=(0{)}^2–0+4$
$=0+4$
$=4$

Thus, $f(0)=4$.

---

Example 95:

Find the range of the function $f(x)=2x^2–5x+3$ for the domain $1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(1)=2(1{)}^2–5(1)+3=2–5+3=0$
$f(2)=2(2{)}^2–5(2)+3=8–10+3=1$
$f(3)=2(3{)}^2–5(3)+3=18–15+3=6$

Thus, the range is $0,1,6$.

---

Example 96:

Determine if the relation $R=(1,2),(2,3),(3,4),(1,5)$ is a function.

Solution:

Step 1: Check if each element in the domain corresponds to exactly one element in the range.

For $x=1$, there are two outputs: $y=2$ and $y=5$. Since one $x$ has more than one $y$, this is not a function.

---

Example 97:

Let $f(x)=x^2+2x–3$. Find $f(2)$.

Solution:

Step 1: Substitute $x=2$ into the function.

$f(2)=(2{)}^2+2(2)–3$
$=4+4–3$
$=5$

Thus, $f(2)=5$.

---

Example 98:

Find the domain and range of the function $f(x)=3x–5$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=3(0)–5=-5$
$f(1)=3(1)–5=-2$
$f(2)=3(2)–5=1$
$f(3)=3(3)–5=4$

Thus, the domain is $0,1,2,3$ and the range is $-5,-2,1,4$.

---

Example 99:

Let $f(x)=4x^2–2x+1$. Find $f(0)$.

Solution:

Step 1: Substitute $x=0$ into the function.

$f(0)=4(0{)}^2–2(0)+1$
$=0+0+1$
$=1$

Thus, $f(0)=1$.

---

Example 100:

Find the range of the function $f(x)=x^2+2x+1$ for the domain $0,1,2,3$.

Solution:

Step 1: Evaluate the function for each value in the domain.

$f(0)=(0{)}^2+2(0)+1=1$
$f(1)=(1{)}^2+2(1)+1=1+2+1=4$
$f(2)=(2{)}^2+2(2)+1=4+4+1=9$
$f(3)=(3{)}^2+2(3)+1=9+6+1=16$

Thus, the range is $1,4,9,16$.