# In-Class Activity 2.1A

## Question 1

Simplify each expression, write without negative exponents.

### a. $\dfrac{x^3y^{-2}z^3}{x^5y^5z^1}$

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<summary> Solution: </summary>

We are given the expression:

$$\frac{x^3y^{-2}z^3}{x^5y^5z^1}$$

### Step 1: Simplify the powers of $x$, $y$, and $z$.

We will apply the quotient rule of exponents, which states:

$$\frac{a^m}{a^n} = a^{m-n}$$

#### Simplifying the powers of $x$:
$$\frac{x^3}{x^5} = x^{3-5} = x^{-2}$$

#### Simplifying the powers of $y$:
$$\frac{y^{-2}}{y^5} = y^{-2-5} = y^{-7}$$

#### Simplifying the powers of $z$:
$$\frac{z^3}{z^1} = z^{3-1} = z^2$$

### Step 2: Write the expression without negative exponents.

Using the property $a^{-n} = \frac{1}{a^n}$, we rewrite $x^{-2}$ and $y^{-7}$ as follows:

$$x^{-2} = \frac{1}{x^2}, \quad y^{-7} = \frac{1}{y^7}$$

Thus, the expression becomes:

$$\frac{z^2}{x^2 y^7}$$

### Final Answer:

$$\frac{z^2}{x^2 y^7}$$

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### b. $\dfrac{x^2y^5z^{-3}}{x^4y^{-3}z^2}$

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<summary> Solution: </summary>

We are given the expression:

$$\frac{x^2y^5z^{-3}}{x^4y^{-3}z^2}$$

### Step 1: Simplify the powers of $x$, $y$, and $z$.

We will apply the quotient rule of exponents, which states:

$$\frac{a^m}{a^n} = a^{m-n}$$

#### Simplifying the powers of $x$:
$$\frac{x^2}{x^4} = x^{2-4} = x^{-2}$$

#### Simplifying the powers of $y$:
$$\frac{y^5}{y^{-3}} = y^{5 - (-3)} = y^{5+3} = y^8$$

#### Simplifying the powers of $z$:
$$\frac{z^{-3}}{z^2} = z^{-3-2} = z^{-5}$$

### Step 2: Write the expression without negative exponents.

Using the property $a^{-n} = \frac{1}{a^n}$, we rewrite $x^{-2}$ and $z^{-5}$ as follows:

$$x^{-2} = \frac{1}{x^2}, \quad z^{-5} = \frac{1}{z^5}$$

Thus, the expression becomes:

$$\frac{y^8}{x^2 z^5}$$

### Final Answer:

$$\frac{y^8}{x^2 z^5}$$

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## Question 2

Factor each polynomial into the product of first-degree factors with integer coefficients.

### a. $x^2-4x-21$

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<summary> Solution: </summary>

We are given the quadratic expression:

$$x^2 - 4x - 21$$

### Step 1: Identify the factors of the constant term and middle coefficient.

We need to find two numbers that multiply to give the constant term, $-21$, and add to give the coefficient of the middle term, $-4$.

- The product of the two numbers must be $-21$.
- The sum of the two numbers must be $-4$.

The factors of $-21$ that satisfy these conditions are $-7$ and $3$, because:

$$-7 \times 3 = -21 \quad \text{and} \quad -7 + 3 = -4$$

### Step 2: Write the factored form.

Now, we can write the quadratic expression as a product of two binomials using these factors:

$$x^2 - 4x - 21 = (x - 7)(x + 3)$$

### Final Answer:

$$x^2 - 4x - 21 = (x - 7)(x + 3)$$

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### b. $x^3+15x^2+50x$

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<summary> Solution: </summary>

We are given the polynomial expression:

$$x^3 + 15x^2 + 50x$$

### Step 1: Factor out the greatest common factor (GCF).

The GCF of all the terms is $x$, so we can factor $x$ out of the expression:

$$x(x^2 + 15x + 50)$$

### Step 2: Factor the quadratic inside the parentheses.

We now focus on factoring the quadratic $x^2 + 15x + 50$. We need to find two numbers that multiply to give the constant term, $50$, and add to give the middle coefficient, $15$.

- The product of the two numbers must be $50$.
- The sum of the two numbers must be $15$.

The factors of $50$ that satisfy these conditions are $5$ and $10$, because:

$$5 \times 10 = 50 \quad \text{and} \quad 5 + 10 = 15$$

### Step 3: Write the fully factored form.

Using these factors, we can write the quadratic expression as:

$$x^2 + 15x + 50 = (x + 5)(x + 10)$$

So, the fully factored form of the original expression is:

$$x(x + 5)(x + 10)$$

### Final Answer:

$$x^3 + 15x^2 + 50x = x(x + 5)(x + 10)$$

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### c. $6x^2-x-1$

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<summary> Solution: </summary>

We are given the quadratic expression:

$$6x^2 - x - 1$$

### Step 1: Set up a multiplication table.

We want to factor this into two binomials. The product of these binomials should match the quadratic expression.

Split $6x^2=(3x)(2x)$ and $-1=(-1)(1)$. These form the margins of the table. Setting up the table:

| | $2x$ | $-1$ |
|------------|------------|------------|
| $3x$ | $6x^2$ | $-3x$ |
| $1$ | $2x$ | $-1$ |

### Step 2: Check the table.

- $3x \cdot 2x = 6x^2$ (this gives the $6x^2$ term)
- $3x \cdot (-1) = -3x$ (this gives part of the middle term)
- $1 \cdot 2x = 2x$ (this gives the other part of the middle term)
- $1 \cdot (-1) = -1$ (this gives the constant term)

### Step 3: Combine the middle terms.

Adding $-3x$ and $2x$ gives $-x$, which matches the middle term of the original expression.

### Final Answer:

So, the factored form is:

$$6x^2 - x - 1 = (3x + 1)(2x - 1)$$

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## Question 3

Use the graph of the function $f$ shown below to estimate the indicated limits and function values.


### a. $\displaystyle \lim_{x \to 2^-}f(x)$

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<summary> Solution: </summary>

The limit as $x$ approaches 2 from the left can be found by looking at the point $(2,1)$ on the graph. Thus the limit is 1.

$\displaystyle \lim_{x \to 2^-}f(x)=1$

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### b. $\displaystyle \lim_{x \to 2^+}f(x)$

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<summary> Solution: </summary>

The limit as $x$ approaches 2 from the right can be found by looking at the point $(2,2)$ on the graph. Thus the limit is 2.

$\displaystyle \lim_{x \to 2^+}f(x)=2$

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### c. $\displaystyle \lim_{x \to 2}f(x)$

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<summary> Solution: </summary>

Since the limits from the left and right are not equal in parts a and b, the limit doesn't exist.

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### d. $f(2.25)$

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<summary> Solution: </summary>

Note that the graph passes through the point $(2.25,2.25)$. Thus $f(2.25)=2.25$.

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### e. $\displaystyle \lim_{x \to 4^-}f(x)$

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<summary> Solution: </summary>

The limit as $x$ approaches 4 from the left can be found by looking at the point $(4,4)$ on the graph. Thus the limit is 4.

$\displaystyle \lim_{x \to 4^-}f(x)=4$

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### f. $\displaystyle \lim_{x \to 4^+}f(x)$

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<summary> Solution: </summary>

The limit as $x$ approaches 4 from the right can be found by looking at the point $(4,4)$ on the graph. Thus the limit is 4.

$\displaystyle \lim_{x \to 4^+}f(x)=4$

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### g. $\displaystyle \lim_{x \to 4} f(x)$

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<summary> Solution: </summary>

Since both the left and right limits are equal ($\displaystyle \lim_{x \to 4^-}f(x)=4$ and $\displaystyle \lim_{x \to 4^+}f(x)=4$), we have:

$$\lim_{x \to 4}f(x)=4$$

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### h. $f(4)$

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<summary> Solution: </summary>

Since there is an open circle at the point $(4,4)$, $f(4)$ doesn't exist.

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## Question 4

Let $$f(x)=\dfrac{x+3}{x^2+3x}$$

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<summary> Simplification (ALWAYS SIMPLIFY FIRST BEFORE TAKING LIMITS!!!!): </summary>

\begin{align}
f(x)&=\dfrac{x+3}{x^2+3x} \\
&=\dfrac{x+3}{x(x+3)} \\
&=\dfrac{1}{x}
\end{align}

$$ \text{Simplified: } f(x)=\dfrac{1}{x}$$
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Find:

### a. $\displaystyle \lim_{x \to -3} f(x)$

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<summary> Solution: </summary>

Plug $x=-3$ into the simplified $f(x)$.

\begin{align}
\lim_{x \to -3} f(x)&=\lim_{x \to -3} \dfrac{1}{x} \\
&=\dfrac{1}{-3} \\
&=-\dfrac{1}{3}
\end{align}

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### b. $\displaystyle \lim_{x \to 0} f(x)$

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<summary> Solution: </summary>

Plug $x=0$ into the simplified $f(x)$.

\begin{align}
\lim_{x \to 0} f(x)&=\lim_{x \to 0} \dfrac{1}{x} \\
&=\dfrac{1}{0} \text{ Error. Test Values Near 0.}
\end{align}

### Testing values near $x = 0$ for the function $f(x) = \frac{1}{x}$ using an $x/y$ table and limit notation

We will examine the behavior of the function $f(x) = \frac{1}{x}$ as $x$ approaches $0$ from both the right ($x \to 0^+$) and the left ($x \to 0^-$). To do this, we will create an $x/y$ table of values near $x = 0$.

### 1. Limit as $x$ approaches $0$ from the right ($x \to 0^+$):

| $x$ | $f(x) = \frac{1}{x}$ |
|-----------|------------------------|
| 0.1 | $f(0.1) \approx 10$ |
| 0.01 | $f(0.01) \approx 100$ |
| 0.001 | $f(0.001) \approx 1000$|

As $x$ gets closer to $0$ from the right (positive values), the function values become larger and larger, approaching $+\infty$. In limit notation:

$$\lim_{x \to 0^+} \frac{1}{x} = +\infty$$

### 2. Limit as $x$ approaches $0$ from the left ($x \to 0^-$):

| $x$ | $f(x) = \frac{1}{x}$ |
|-----------|------------------------|
| -0.1 | $f(-0.1) \approx -10$ |
| -0.01 | $f(-0.01) \approx -100$|
| -0.001 | $f(-0.001) \approx -1000$|

As $x$ gets closer to $0$ from the left (negative values), the function values become more and more negative, approaching $-\infty$. In limit notation:

$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$

### Conclusion:

- As $x \to 0^+$, $f(x)$ increases without bound:
$$\lim_{x \to 0^+} \frac{1}{x} = +\infty$$

- As $x \to 0^-$, $f(x)$ decreases without bound:
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$

Thus, the function $f(x) = \frac{1}{x}$ has a vertical asymptote at $x = 0$, with different behavior on either side of $0$.

Since the limits from the left and right are different, $\displaystyle \lim_{x \to 0}f(x)$ doesn't exist.

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### c. $\displaystyle \lim_{x \to 3} f(x)$

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<summary> Solution: </summary>

Plug $x=3$ into the simplified $f(x)$.

\begin{align}
\lim_{x \to 3} f(x)&=\lim_{x \to 3} \dfrac{1}{x} \\
&=\dfrac{1}{3}
\end{align}

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