---
# System prepended metadata

title: In-Class Activity 2.1B
tags: [in-class-activity]

---

# In-Class Activity 2.1B

## Question 1

Simplify each expression, write without negative or fractional exponents.

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


::: spoiler

<summary> Solution: </summary>

We are given the expression:

$$\frac{x^3 (y^2z^3)^2}{\sqrt{x^5y^5} z^1}$$

### Step 1: Simplify the numerator

First, simplify the expression in the numerator:

$$(y^2z^3)^2$$

This means that both $y^2$ and $z^3$ are raised to the power of 2. Using the exponent rule $(a^m)^n = a^{m \cdot n}$, we get:

$$y^{2 \cdot 2}z^{3 \cdot 2} = y^4z^6$$

Now, substitute this back into the numerator:

$$x^3 \cdot y^4 \cdot z^6$$

### Step 2: Simplify the denominator

In the denominator, we have:

$$\sqrt{x^5y^5} \cdot z^1$$

The square root of $x^5y^5$ can be simplified as:

$$\sqrt{x^5y^5} = x^{5/2}y^{5/2}$$

So the denominator becomes:

$$x^{5/2}y^{5/2} \cdot z^1$$

### Step 3: Combine the expression

Now, combine the numerator and denominator:

$$\frac{x^3y^4z^6}{x^{5/2}y^{5/2}z^1}$$

### Step 4: Apply exponent rules

We will now simplify the expression using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

#### Simplifying $x$ terms:
$$\frac{x^3}{x^{5/2}} = x^{3 - 5/2} = x^{6/2 - 5/2} = x^{1/2}$$

#### Simplifying $y$ terms:
$$\frac{y^4}{y^{5/2}} = y^{4 - 5/2} = y^{8/2 - 5/2} = y^{3/2}$$

#### Simplifying $z$ terms:
$$\frac{z^6}{z^1} = z^{6 - 1} = z^5$$

### Step 5: Final simplified expression

The fully simplified expression is:

$$x^{1/2}y^{3/2}z^5$$

Writing without fractional exponents, you can write the final answer as:

$$\sqrt{xy^3}  z^5$$


:::

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

::: spoiler

<summary> Solution: </summary>

We are given the expression:

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

### Step 1: Simplify the square root in the numerator

The square root can be rewritten as:

$$\sqrt{y^5z^{-3}} = (y^5z^{-3})^{1/2} = y^{5/2}z^{-3/2}$$

Now, substitute this back into the numerator:

$$x^2 \cdot y^{5/2} \cdot z^{-3/2}$$

### Step 2: Simplify the denominator

We now simplify the denominator:

$$(x^4y^{-3})^3z^2$$

Apply the exponent rule $(a^m)^n = a^{m \cdot n}$:

$$x^{4 \cdot 3}y^{-3 \cdot 3} = x^{12}y^{-9}$$

Now the denominator becomes:

$$x^{12}y^{-9}z^2$$

### Step 3: Combine the expression

Now, combine the simplified numerator and denominator:

$$\frac{x^2y^{5/2}z^{-3/2}}{x^{12}y^{-9}z^2}$$

### Step 4: Apply exponent rules

We will now simplify the expression using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

#### Simplifying $x$ terms:
$$\frac{x^2}{x^{12}} = x^{2 - 12} = x^{-10}$$

#### Simplifying $y$ terms:
$$\frac{y^{5/2}}{y^{-9}} = y^{5/2 - (-9)} = y^{5/2 + 9} = y^{5/2 + 18/2} = y^{23/2}$$

#### Simplifying $z$ terms:
$$\frac{z^{-3/2}}{z^2} = z^{-3/2 - 2} = z^{-3/2 - 4/2} = z^{-7/2}$$

### Step 5: Convert fractional exponents to radicals

- For $x^{-10}$, this is equivalent to $\dfrac{1}{x^{10}}$.
- For $y^{23/2}$, this is equivalent to $\sqrt{y^{23}}$.
- For $z^{-7/2}$, this is equivalent to $\frac{1}{\sqrt{z^7}}$.

### Step 6: Final simplified expression

Now, write the final expression without fractional exponents:

$$\dfrac{ \sqrt{y^{23}}}{x^{10}\sqrt{z^7}}$$



:::

## Question 2

Let $$f(x)=2x^2-3x+5$$ $$g(x)=3x^2+5x-1$$

Simplify:

### a. $f(x+h)-f(x)$

::: spoiler

<summary> Solution: </summary>

Let's begin by simplifying $f(x+h) - f(x)$, where $f(x) = 2x^2 - 3x + 5$.

### Step 1: Find $f(x+h)$

Substitute $x+h$ into $f(x)$:

$$
f(x+h) = 2(x+h)^2 - 3(x+h) + 5
$$

Now expand the terms:

$$
f(x+h) = 2(x^2 + 2xh + h^2) - 3(x + h) + 5
$$

$$
f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h + 5
$$

### Step 2: Simplify $f(x+h) - f(x)$

Now subtract $f(x)$ from $f(x+h)$:

$$
f(x+h) - f(x) = \left(2x^2 + 4xh + 2h^2 - 3x - 3h + 5\right) - \left(2x^2 - 3x + 5\right)
$$

Distribute the negative sign:

$$
f(x+h) - f(x) = \left(2x^2 + 4xh + 2h^2 - 3x - 3h + 5\right) - 2x^2 + 3x - 5
$$

### Step 3: Cancel out like terms

Cancel the terms that appear in both $f(x+h)$ and $f(x)$:

$$
f(x+h) - f(x) = 4xh + 2h^2 - 3h
$$

### Final Answer:

$$
f(x+h) - f(x) = 4xh + 2h^2 - 3h
$$




:::

### b. $g(x+h)-g(x-h)$

::: spoiler

<summary> Solution: </summary>

Let's simplify $g(x+h) - g(x-h)$, where $g(x) = 3x^2 + 5x - 1$.

### Step 1: Find $g(x+h)$

Substitute $x+h$ into $g(x)$:

$$
g(x+h) = 3(x+h)^2 + 5(x+h) - 1
$$

Now expand the terms:

$$
g(x+h) = 3(x^2 + 2xh + h^2) + 5(x + h) - 1
$$

$$
g(x+h) = 3x^2 + 6xh + 3h^2 + 5x + 5h - 1
$$

### Step 2: Find $g(x-h)$

Substitute $x-h$ into $g(x)$:

$$
g(x-h) = 3(x-h)^2 + 5(x-h) - 1
$$

Now expand the terms:

$$
g(x-h) = 3(x^2 - 2xh + h^2) + 5(x - h) - 1
$$

$$
g(x-h) = 3x^2 - 6xh + 3h^2 + 5x - 5h - 1
$$

### Step 3: Simplify $g(x+h) - g(x-h)$

Now subtract $g(x-h)$ from $g(x+h)$:

$$
g(x+h) - g(x-h) = \left(3x^2 + 6xh + 3h^2 + 5x + 5h - 1\right) - \left(3x^2 - 6xh + 3h^2 + 5x - 5h - 1\right)
$$

Distribute the negative sign:

$$
g(x+h) - g(x-h) = \left(3x^2 + 6xh + 3h^2 + 5x + 5h - 1\right) - 3x^2 + 6xh - 3h^2 - 5x + 5h + 1
$$

### Step 4: Cancel out like terms

Cancel the terms that appear in both $g(x+h)$ and $g(x-h)$:

$$
g(x+h) - g(x-h) = 6xh + 5h + 6xh + 5h
$$

Simplify the expression:

$$
g(x+h) - g(x-h) = 12xh + 10h
$$

### Final Answer:

$$
g(x+h) - g(x-h) = 12xh + 10h
$$


:::


## Question 3

Let $$f(x)=\dfrac{2x^2-3x-2}{x^2+x-6}$$

::: spoiler

<summary> Simplification (ALWAYS SIMPLIFY BEFORE PLUGGING IN LIMITS!!!!): </summary>

### Simplification of $\dfrac{2x^2 - 3x - 2}{x^2 + x - 6}$

#### Step 1: Factor the numerator and the denominator

1. **Factor the numerator** $2x^2 - 3x - 2$:

   The factored form is:

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

   Verify by expanding:

   $$
   (2x + 1)(x - 2)
   $$

   |        | $x$    | $-2$   |
   |--------|--------|--------|
   | $2x$   | $2x \cdot x = 2x^2$ | $2x \cdot (-2) = -4x$ |
   | $1$    | $1 \cdot x = x$ | $1 \cdot (-2) = -2$   |

   Combine the terms:

   $$
   2x^2 - 4x + x - 2 = 2x^2 - 3x - 2
   $$

2. **Factor the denominator** $x^2 + x - 6$:

   The factored form is:

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

   Verify by expanding:

   $$
   (x + 3)(x - 2)
   $$

   |        | $x$    | $-2$   |
   |--------|--------|--------|
   | $x$    | $x \cdot x = x^2$ | $x \cdot (-2) = -2x$ |
   | $3$    | $3 \cdot x = 3x$  | $3 \cdot (-2) = -6$   |

   Combine the terms:

   $$
   x^2 - 2x + 3x - 6 = x^2 + x - 6
   $$

#### Step 2: Simplify the expression

Now, the expression becomes:

$$
\dfrac{(2x + 1)(x - 2)}{(x + 3)(x - 2)}
$$

Since $(x - 2)$ appears in both the numerator and the denominator, cancel it out:

$$
\dfrac{2x + 1}{x + 3}
$$

#### Final Simplified Form:

The simplified function is:

$$
f(x)=\dfrac{2x + 1}{x + 3}
$$


:::


Find:

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

::: spoiler

<summary> Solution: </summary>

\begin{align}
\lim_{x \to 2} f(x)&=\lim_{x \to 2} \dfrac{2x + 1}{x + 3} \\
&=\dfrac{2(2)+1}{2+3} \\
&=\dfrac{5}{5} \\
&=1
\end{align}



:::

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

::: spoiler

<summary> Solution: </summary>

\begin{align}
\lim_{x \to 0} f(x)&=\lim_{x \to 0} \dfrac{2x + 1}{x + 3} \\
&=\dfrac{2(0)+1}{0+3} \\
&=\dfrac{1}{3}
\end{align}

:::

### c. $\displaystyle \lim_{x \to 1}f(x)$

::: spoiler

<summary> Solution: </summary>

\begin{align}
\lim_{x \to 1} f(x)&=\lim_{x \to 1} \dfrac{2x + 1}{x + 3} \\
&=\dfrac{2(1)+1}{1+3} \\
&=\dfrac{3}{4}
\end{align}

:::


## Question 4

Let $$f(x)=-4x+13$$

Find:

 $$\displaystyle \lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}$$

::: spoiler

<summary> Solution: </summary>

### Given Function:
$$f(x) = -4x + 13$$

We need to find:

$$
\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}
$$

#### Step 1: Find $f(2+h)$ and $f(2)$

1. **Find $f(2+h)$**:

   Using the function $f(x) = -4x + 13$, substitute $x = 2+h$:

   $$
   f(2+h) = -4(2+h) + 13
   $$

   Expand:

   $$
   f(2+h) = -4 \cdot 2 - 4 \cdot h + 13 = -8 - 4h + 13 = 5 - 4h
   $$

2. **Find $f(2)$**:

   Substitute $x = 2$ into the function:

   $$
   f(2) = -4(2) + 13 = -8 + 13 = 5
   $$

#### Step 2: Plug into the limit expression

Now, substitute $f(2+h) = 5 - 4h$ and $f(2) = 5$ into the limit expression:

$$
\lim_{h \to 0} \dfrac{(5 - 4h) - 5}{h}
$$

Simplify the numerator:

$$
\lim_{h \to 0} \dfrac{5 - 4h - 5}{h} = \lim_{h \to 0} \dfrac{-4h}{h}
$$

Cancel the $h$ terms:

$$
\lim_{h \to 0} -4 = -4
$$

#### Final Answer:

$$
\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h} = -4
$$


:::

