---
# System prepended metadata

title: 1.2 lecture notes
tags: ['1.2']

---

## JIT 5. Number line inequality


![image](https://hackmd.io/_uploads/Bk-YqDCuke.png)


### Examples



True or false?




| Inequality  | Verbal Translation                | True/False |
|------------|----------------------------------|------------|
| 5 > -4     | 5 is greater than -4            | True       |
| -2 ≤ -7    | -2 is less than or equal to -7  | False      |
| 2 < -10    | 2 is less than -10              | False      |
| 8 ≥ -9     | 8 is greater than or equal to -9| True       |
| -5 ≥ -7    | -5 is greater than or equal to -7 | True    |
| -3 ≤ 4     | -3 is less than or equal to 4   | True       |

## JIT 6  Interval Notation; sets of real numbers


### Figure 2:

![image](https://hackmd.io/_uploads/r1_n9wCu1e.png)

 

Inequality: $-1 \leq x \leq 3$

Interval notation: $[-1,3]$. (Include $-1,3$).


### Figure 3:

![image](https://hackmd.io/_uploads/r1TIov0dkx.png)


 

Inequality: $-1 < x < 3$

Interval notation: $(-1,3)$. (Exclude $-1,3$).

### Figure 4:

![image](https://hackmd.io/_uploads/HkLPoPRdke.png)


 

Inequality: $-1 \leq x < 3$

Interval notation: $[-1,3)$. (Include $-1$, exclude $3$).



### Figure 5:

![image](https://hackmd.io/_uploads/HyUrnwAdkl.png)



 

Inequality: $-1 < x \leq 3$

Interval notation: $[-1,3)$. (Exclude $-1$, include $3$).

### Figure 6:

![image](https://hackmd.io/_uploads/BkJUnwC_1l.png)



 

Inequality: $-1 <x$ or $x >-1$

Interval notation: $(-1,\infty)$. (Exclude $-1$, always exclude $\infty$).

### Figure 7:

![image](https://hackmd.io/_uploads/H18L3DRdkl.png)



 

Inequality: $-1 \leq x$ or $x \geq -1$

Interval notation: $[-1,\infty)$. (Include $-1$, always exclude $\infty$).

### Figure 8:

![image](https://hackmd.io/_uploads/B1zdnvRd1g.png)



 

Inequality: $x<3$ or $3>x$

Interval notation: $(-\infty,3)$. (Always exclude $-\infty$, exclude $3$)


### Figure 9:

![image](https://hackmd.io/_uploads/S13u3DAd1x.png)



 

Inequality: $x\leq 3$ or $3 \geq x$

Interval notation: $(-\infty,3]$. (Always exclude $-\infty$, include $3$)


- In $(a, b)$ or $[a, b]$ notation, $a$ is always less than $b$.
- The $\infty$ symbol is only on the right.
- The $-\infty$ symbol is only on the left.

### Examples
Draw each of the following on a number line and give the inequality. If the notation is wrong, state why.

1. $[2, 5]$  
   Inequality: $2 \leq x \leq 5$  
   *Correct notation.* 2 is less than 5.

2. $(2, 7]$  
   Inequality: $2 < x \leq 7$  
   *Correct notation.* 2 is less than 7.

3. $[8, 5]$  
   *Incorrect notation: Lower bound (8) is greater than upper bound (5).* 8 is not less than 5.

4. $(-8, 1]$  
   Inequality: $-8 < x \leq 1$  
   *Correct notation.* -8 is less than 1.

5. $(-\infty, 4]$  
   Inequality: $x \leq 4$  
   *Correct notation.* $-\infty$ is less than 4.

6. $(3, \infty)$  
   Inequality: $x > 3$  
   *Correct notation.* $3$ is less than $\infty$.

7. $(1, 7)$  
   Inequality: $1 < x < 7$  
   *Correct notation.* 1 is less than 7.

8. $(3, -\infty)$  
   *Incorrect notation: The lower bound (3) must be smaller than the upper bound (-∞).* Should be $(-\infty, 3)$.

9. $[-2, \infty)$  
   Inequality: $-2 \leq x$  
   *Correct notation.* -2 is less than $\infty$.

10. $(\infty, 6)$  
   *Incorrect notation: Infinity cannot be the starting value.* Should be $(-\infty, 6)$.

11. $[1, -6]$  
   *Incorrect notation: The lower bound (1) must be smaller than the upper bound (-6).* Should be $[-6, 1]$.

12. $(8, 0)$  
   *Incorrect notation: The lower bound (8) must be smaller than the upper bound (0).* Should be $(0, 8)$.


## 1.2 Functions and graphs.
- A **function** is a correspondence between a set of inputs, called the **domain**, and a set of outputs, called the **range** such that every input in the domain corresponds to exactly one member of the range.


- Put another way, every input produces exactly one output.

| domain (inputs) | range (outputs) |
|----------------|----------------|
| 1              | 3              |
| 2              | 5              |
| 3              | 7              |
| 3              | 8              |
| 4              | 3              |
| 4              | 7              |
| 5              | 6              |


- Not a function because the input 3 has two outputs 7 and 8. Also since 4 has two outputs 3 and 7.

---

| domain (inputs) | range (outputs) |
|----------------|----------------|
| 1              | 3              |
| 2              | 5              |
| 3              | 7              |
| 4              | 3              |
| 5              | 6              |

- This is a function since every input has only one output.

- The fact that 3 appears twice in the range is not a problem.


---

### **Example.** Is this a function? Explain.

| domain | range |
|--------|-------|
| 1      | 3     |
| 2      | 3     |
| 3      | 7     |
| 4      | 3     |
| 5      | 3     |
| 6      | 7     |
| 7      | 3     |
| 8      | 7     |


Answer: Yes, since every input goes to one output.


---

### **Example.** Is this a function? Explain.

| domain | range |
|--------|-------|
| 1      | 3     |
| 2      | 3     |
| 3      | 7     |
| 4      | 3     |
| 4      | 7     |
| 6      | 7     |
| 7      | 3     |
| 8      | 7     |


**Answer**: No, since the input 4 goes to the output 3 and 7.

### **Example.** Is this a function? Explain. (The domain and range don’t have to be numbers.)

| domain | range |
|--------|-------|
| a      | 3     |
| b      | 5     |
| c      | 9     |
| d      | 11    |
| b      | 13    |
| c      | 15    |
| f      | 17    |
| g      | 19    |

**Answer**: No, since the input $c$ has two outputs 9 and 15.

---

- It is more compact to express the correspondence as ordered pairs \((x, y)\) where \(x\) is in the domain and \(y\) is in the range. For example, the function

### Ordered pairs.

| domain | range |
|--------|-------|
| 1      | 3     |
| 2      | 3     |
| 3      | 7     |
| 4      | 3     |
| 6      | 7     |
| 7      | 3     |
| 8      | 7     |

can be written as $\{(1,3), (2,3), (3,7), (4,3), (6,7), (7,3), (8,7)\}$.


Note that the above is a function since every input has exactly one output.


### **Example.** Is the following a function? Identify the domain and range.

$$\{(1,3), (2,3), (3,7), (4,3), (6,7), (6,9), (7,3), (8,7), (8,9)\}$$

**Answer**: No, since one input 6 goes to two outputs 7 and 9.

## Using arrows between domain and range.

- On the left we have inputs and on the right we have outputs:

**Question:** Is this a function?

![image](https://hackmd.io/_uploads/rkvPd-Ztkl.png)

- This says input 1 goes to 2, 3, and 4.

- Input 2 goes to 3 and 4.

- Input 3 goes to 1 and 3.

- Input 4 goes to 2 and 4.


**Answer:** No, one input goes to more than one output.


---

**Question:** Is this a function?

![image](https://hackmd.io/_uploads/BysO_WbY1l.png)



**Answer:** No, one input 2 goes to more than one output 3 and 4.


---

**Question:** Is this a function?


![image](https://hackmd.io/_uploads/HyGYu-WYJl.png)



**Answer:** Yes, every input goes to exactly one output. The fact that multiple inputs 2 and 3 go to single output 1 does not matter.

---

**Question:** Is this a function?

![image](https://hackmd.io/_uploads/BktY_bbt1x.png)


**Answer:** No, since one input 3 goes to two outputs 1 and 3.

---

**Question:** Is this a function?

![image](https://hackmd.io/_uploads/B175OZWK1x.png)



**Answer:** No, since one input 3 goes to two outputs 2 and 3.

---


**Question:** Is this a function?

![image](https://hackmd.io/_uploads/BJj5_-Wtkx.png)


**Answer:** Yes, every input goes to exactly one output.

---


**Question:** Is this a function?

![image](https://hackmd.io/_uploads/B1bjdWZtyx.png)


**Answer:** No, since one input 2 goes to two outputs 3 and 4.


---


### Introduction
We will explore how to evaluate a linear function $f(x) = 2x + 5$ by plugging in different values of $x$. This process shows how the input in the domain corresponds to an output in the range.

---

#### **Example 1a**: Find $f(3)$ for $y = f(x) = 2x + 5$.

\begin{align}
f(x) &= 2x + 5 \\
f(3) &= 2(3) + 5 \\
&= 6 + 5 \\
&= 11
\end{align}

Thus, $f(3) = 11$.

---

#### **Example 1b**: Find $f(-7)$ for $y = f(x) = 2x + 5$.

\begin{align}
f(x) &= 2x + 5 \\
f(-7) &= 2(-7) + 5 \\
&= -14 + 5 \\
&= -9
\end{align}

Thus, $f(-7) = -9$.

---

#### **Example 1c**: Find $f(-5)$ for $y = f(x) = 2x + 5$.

\begin{align}
f(x) &= 2x + 5 \\
f(-5) &= 2(-5) + 5 \\
&= -10 + 5 \\
&= -5
\end{align}

Thus, $f(-5) = -5$.

---

#### **Example 1d**: Find $f\left( \frac{3}{7} \right)$ for $y = f(x) = 2x + 5$.

\begin{align}
f(x) &= 2x + 5 \\
f\left( \frac{3}{7} \right) &= 2 \left( \frac{3}{7} \right) + 5 \\
&= \frac{6}{7} + 5 \\
&= \frac{6}{7} + \frac{35}{7} \quad \text{(convert 5 to } \frac{35}{7}\text{)} \\
&= \frac{41}{7}
\end{align}

---

Thus, $f\left( \frac{3}{7} \right) = \frac{41}{7}$.

---

#### **Example 1e**: Find $f\left( -\frac{2}{5} \right)$ for $y = f(x) = 2x + 5$.

\begin{align}
f(x) &= 2x + 5 \\
f\left( -\frac{2}{5} \right) &= 2 \left( -\frac{2}{5} \right) + 5 \\
&= -\frac{4}{5} + 5 \\
&= -\frac{4}{5} + \frac{25}{5} \quad \text{(convert 5 to } \frac{25}{5}\text{)} \\
&= \frac{21}{5}
\end{align}

---

Thus, $f\left( -\frac{2}{5} \right) = \frac{21}{5}$.

---

#### **Example 2a.** For $y = g(x) = x^2$, find:

(a) $g(3) = 3^2 = 3 \cdot 3=9$

---

#### **Example 2b.** For $y = g(x) = x^2$, find:

(b) $g\left( -\frac{2}{7} \right)$

\begin{align}
g\left( -\frac{2}{7} \right) &= \left( -\frac{2}{7} \right)^2 \\
&= \left( -\frac{2}{7} \right) \left( -\frac{2}{7} \right) \\
&= \frac{(-2) \cdot (-2)}{7 \cdot 7} \\
&= \frac{4}{49}
\end{align}

---

#### **Example 2c .** For $y = g(x) = x^2$, find:

(c) $g(3t)$

\begin{align}
g(3t)&= (3t)^2  \\
&=(3t)(3t) \\
&= (3 \cdot 3)(t \cdot t)  \\
&= 9t^2
\end{align}

---

#### **Example 2d.** For $y = g(x) = x^2$, find:

(d) $g\left( -\frac{6t}{5} \right)$

\begin{align} 
g\left( -\frac{6t}{5} \right)&= \left( -\frac{6t}{5} \right)^2 \\
&= \left( -\frac{6t}{5} \right)\left( -\frac{6t}{5} \right) \\
&= \frac{36t^2}{25}
\end{align}

---

#### **Example 2e.** For $y = g(x) = x^2$, find:

(e) $g(s + t)$

\begin{align}
g(s + t) &= (s + t)^2 \\
&=(s+t)(s+t) 
\end{align}

To expand $(s + t)(s + t)$, we use the distributive property.

---

To expand $(s + t)(s + t)$, we use the distributive property. Here’s the multiplication table:

|        | $s$   | $t$   |
|--------|-------|-------|
| **$s$**| $s^2$ | $st$  |
| **$t$**| $ts$  | $t^2$ |

---

Now, sum the products:

$(s + t)(s + t) = s^2 + st + ts + t^2$

Since $st = ts$, this simplifies to:

$g(s+t)=(s + t)^2 = s^2 + 2st + t^2$

---

**Example 3a.** For $y = h(x) = \frac{x + 2}{x - 1}$, find:


(a) $h(3)$

\begin{align}
h(3) &= \frac{3 + 2}{3 - 1} \\
&= \frac{5}{2}
\end{align}

Thus, $h(3) = \frac{5}{2}$.

---



**Example 3b.** For $y = h(x) = \frac{x + 2}{x - 1}$, find:

(b) $h\left( \frac{2}{3} \right)$

\begin{align}
h\left( \frac{2}{3} \right) &= \frac{\frac{2}{3} + 2}{\frac{2}{3} - 1} \\
&= \frac{\frac{2}{3} + \frac{6}{3}}{\frac{2}{3} - \frac{3}{3}} \\
&= \frac{\frac{8}{3}}{-\frac{1}{3}} \\
&= -8
\end{align}

---

Thus, $h\left( \frac{2}{3} \right) = -8$.

---



**Example 3c.** For $y = h(x) = \frac{x + 2}{x - 1}$, find:


(c) $h(-5)$

\begin{align}
h(-5) &= \frac{-5 + 2}{-5 - 1} \\
&= \frac{-3}{-6} \\
&= \frac{1}{2}
\end{align}

Thus, $h(-5) = \frac{1}{2}$.

---


**Example 3d.** For $y = h(x) = \frac{x + 2}{x - 1}$, find:
(d) $h(-9)$

\begin{align}
h(-9) &= \frac{-9 + 2}{-9 - 1} \\
&= \frac{-7}{-10} \\
&= \frac{7}{10}
\end{align}

Thus, $h(-9) = \frac{7}{10}$.

---

**Example 3d.** For $y = h(x) = \frac{x + 2}{x - 1}$, find $h\left( \frac{3}{5} \right)$:

\begin{align}
h\left( \frac{3}{5} \right) &= \frac{\frac{3}{5} + 2}{\frac{3}{5} - 1} \\
&= \frac{\frac{3}{5} + \frac{10}{5}}{\frac{3}{5} - \frac{5}{5}} \\
&= \frac{\frac{13}{5}}{-\frac{2}{5}} \\
&= -\frac{13}{2}
\end{align}

---

Thus, $h\left( \frac{3}{5} \right) = -\frac{13}{2}$.

---

(f) What about $h(1)$?

For $x = 1$, we have:

\begin{align}
h(1) &= \frac{1 + 2}{1 - 1} \\
&= \frac{3}{0}
\end{align}

Since division by zero is undefined, $h(1)$ is undefined.

---

Thus, $h(1)$ is undefined.

---

**Example 4a.** For $y = g(x) = \sqrt{1 + x}$, find:

(a) $g(3)$

\begin{align}
g(3) &= \sqrt{1 + 3} \\
&= \sqrt{4} \\
&= 2
\end{align}

Thus, $g(3) = 2$.

---

**Example 4b.** For $y = g(x) = \sqrt{1 + x}$, find $g\left( \frac{1}{4} \right)$:

\begin{align}
g\left( \frac{1}{4} \right) &= \sqrt{1 + \frac{1}{4}} \\
&= \sqrt{\frac{4}{4} + \frac{1}{4}} \\
&= \sqrt{\frac{5}{4}} \\
&= \frac{\sqrt{5}}{2}
\end{align}

Thus, $g\left( \frac{1}{4} \right) = \frac{\sqrt{5}}{2}$.

---

**Example 4c.** For $y = g(x) = \sqrt{1 + x}$, find:

(c) $g(0)$

\begin{align}
g(0) &= \sqrt{1 + 0} \\
&= \sqrt{1} \\
&= 1
\end{align}

Thus, $g(0) = 1$.

---

**Example 4d.** For $y = g(x) = \sqrt{1 + x}$, find:

(d) $g(8)$

\begin{align}
g(8) &= \sqrt{1 + 8} \\
&= \sqrt{9} \\
&= 3
\end{align}

Thus, $g(8) = 3$.

---

**Example 4e.** For $y = g(x) = \sqrt{1 + x}$, find:

(e) $g\left( -\frac{5}{9} \right)$

\begin{align}
g\left( -\frac{5}{9} \right) &= \sqrt{1 - \frac{5}{9}} \\
&= \sqrt{\frac{9}{9} - \frac{5}{9}} \\
&= \sqrt{\frac{4}{9}} \\
&= \frac{2}{3}
\end{align}

Thus, $g\left( -\frac{5}{9} \right) = \frac{2}{3}$.

---

**Example 4f.** For $y = g(x) = \sqrt{1 + x}$, find:

(f) $g(4)$

\begin{align}
g(4) &= \sqrt{1 + 4} \\
&= \sqrt{5}
\end{align}

Thus, $g(4) = \sqrt{5}$.

---

**What can we say about $g(-7)$?**

For $g(x) = \sqrt{1 + x}$, if $x = -7$:

\begin{align}
g(-7) &= \sqrt{1 - 7} \\
&= \sqrt{-6}
\end{align}

Since the square root of a negative number is not defined in the real number system, $g(-7)$ is undefined in the real numbers.

Thus, $g(-7)$ is undefined in the real number system.

---




**Graph of a function**

- The graph of $y = f(x)$ is the set of points $(x, f(x))$ where $x$ is in the domain of $f(x)$.

**Example**: Graph $y = f(x) = 2x - 2$ by plugging in values for $x$ and connecting the points.

---

| **$x$**  | **$y = f(x) = 2x - 2$** |
|:--------:|:-----------------------:|
| $1$      | $2(1) - 2 = 0$          |
| $2$      | $2(2) - 2 = 2$          |
| $3$      | $2(3) - 2 = 4$          |
| $4$      | $2(4) - 2 = 6$          |
| $0$      | $2(0) - 2 = -2$         |
| $-1$     | $2(-1) - 2 = -4$        |
| $-2$     | $2(-2) - 2 = -6$        |
| $5$      | $2(5) - 2 = 8$          |

---

Now, plot the points $(1,0)$, $(2,2)$, $(3,4)$, $(4,6)$, $(0,-2)$, $(-1,-4)$, $(-2,-6)$, and $(5,8)$ on a coordinate plane and connect them to visualize the line for the function $y = 2x - 2$.

---

![image](https://hackmd.io/_uploads/SyTFi-WK1x.png)



---


## Steps to Graph $y = g(x) = x^2 - 5$

1. Choose a range of $x$ values.
2. Plug each $x$ value into the function $g(x) = x^2 - 5$.
3. Record the result as the corresponding $y$ value.
4. Fill out the table with the $x$ and $y$ values.
5. Plot the points on a graph and connect them to visualize the function.

---

### Example:

We will choose $x = 0, 1, 2, 3, 4, -1, -2$ and plug them into the function:


---

| $x$     | $y = g(x) = x^2 - 5$       |
|---------|----------------------------|
| 0       | $0^2 - 5 = -5$             |
| 1       | $1^2 - 5 = -4$             |
| 2       | $2^2 - 5 = -1$             |
| 3       | $3^2 - 5 = 4$              |
| 4       | $4^2 - 5 = 11$             |
| -1      | $(-1)^2 - 5 = -4$          |
| -2      | $(-2)^2 - 5 = -1$          |


---

### Graph:

Now, plot these points and connect them to form the graph of $y = g(x) = x^2 - 5$.


![image](https://hackmd.io/_uploads/BJbynZZtkl.png)


---

### Reading Graphs


**Question** : Using the graph below, find $f(0)$, $f(1)$, $f(2)$, $f(3)$, $f(-1)$, and $f(-2)$.
![image](https://hackmd.io/_uploads/rkwV2ZZFyg.png)


---

### Reading Graphs:

Based on the graph, the values of $f(x)$ are:


| Ordered Pair (x, f(x)) | Function Notation |
|------------------------|------------------|
| (0, 2)                | $f(0) = 2$         |
| (1, 3)                | $f(1) = 3$         |
| (-1, 1)               | $f(-1) = 1$        |
| (2, -2)               | $f(2) = -2$        |
| (3, 2)                | $f(3) = 2$         |
| (-2, 2)               | $f(-2) = 2$        |

