# In-Class Activity 2.4

## Question 1

Factor top and bottom completely, then cancel.

### a. $\dfrac{6x^3+6x^2-12x}{3x^2-3x}$

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<summary> Example: </summary>

Simplify by factoring top and bottom, and then cancelling. $\dfrac{5 x^3 + 5 x^2 - 30 x}{2x^2-4x}$

The top factors as

\begin{align}
5x^3+5x^2-30x&=5x(x^2+x-6) \\
&=5x(x+3)(x-2)
\end{align}

The bottom factors as

\begin{align}
2x^2-4x&=2x(x-2)
\end{align}

Thus the fraction factors as the following:

\begin{align}
\dfrac{5 x^3 + 5 x^2 - 30 x}{2x^2-4x}&=\dfrac{5x(x+3)(x-2)}{2x(x-2)} \\
&=\dfrac{5(x+3)}{2}
\end{align}

Note the $(x-2)$ and $x$ canceled from top and bottom.

#### Final Answer:

$$\dfrac{5(x+3)}{2}$$

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<summary> Solution: </summary>

Simplify by factoring top and bottom, and then cancelling. $\dfrac{6x^3+6x^2-12x}{3x^2-3x}$

The top factors as

\begin{align}
6x^3+6x^2-12x&=6x(x^2+x-2) \\
&=6x(x+2)(x-1)
\end{align}

The bottom factors as

\begin{align}
3x^2-3x&=3x(x-1)
\end{align}

Thus the fraction factors as the following:

\begin{align}
\dfrac{6x^3+6x^2-12x}{3x^2-3x}&=\dfrac{6x(x+2)(x-1)}{3x(x-1)} \\
&=2(x+2)
\end{align}

Note the $(x-1)$ and $x$ and 3 canceled from top and bottom.

#### Final Answer:

$$2(x+2)$$

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### b. $\dfrac{2x^3+10x^2-12x}{x^3-2x^2-3x}$

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<summary> Example: </summary>

Simplify by factoring top and bottom, and then cancelling. $\dfrac{3x^3 + 9x^2 - 12x}{x^3 - 3x^2 - 4x}
$

The top factors as

\begin{align}
3x^3 + 9x^2 - 12x&=3x(x^2+3x-4) \\
&=3x(x+4)(x-1)
\end{align}

The bottom factors as

\begin{align}
x^3 - 3x^2 - 4x&=x(x^2-3x-4) \\
&=x(x-4)(x+1)
\end{align}

Thus the fraction factors as the following:

\begin{align}
\dfrac{3x^3 + 9x^2 - 12x}{x^3 - 3x^2 - 4x}&=\dfrac{3x(x+4)(x-1)}{x(x-4)(x+1)} \\
&=\dfrac{3(x+4)(x-1)}{(x-4)(x+1)}
\end{align}

Note only the $x$ canceled from top and bottom.

#### Final Answer:

$$\dfrac{3(x+4)(x-1)}{(x-4)(x+1)}$$

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<summary> Solution: </summary>

Simplify by factoring top and bottom, and then cancelling. $\dfrac{2x^3+10x^2-12x}{x^3-2x^2-3x}$

The top factors as

\begin{align}
2x^3+10x^2-12x&=2x(x^2+5x-6) \\
&=2x(x+6)(x-1)
\end{align}

The bottom factors as

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

Thus the fraction factors as the following:

\begin{align}
\dfrac{3x^3 + 9x^2 - 12x}{x^3 - 3x^2 - 4x}&=\dfrac{2x(x+6)(x-1)}{x(x-3)(x+1)} \\
&=\dfrac{2(x+6)(x-1)}{(x-3)(x+1)}
\end{align}

Note only the $x$ canceled from top and bottom.

#### Final Answer:

$$\dfrac{2(x+6)(x-1)}{(x-3)(x+1)}$$

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## Question 2

Substitute, expand, and combine terms.

### a. $f(x)=2x^2+2x+4$ ; $f(x+h)-f(x)$

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<summary> Example: </summary>

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

First we find and simplify $f(x+h)$:

\begin{align}
f(x+h)&=3(x+h)^2+4(x+h)+5 \\
&=3(x+h)(x+h)+4(x+h)+5 \\
&=3(x^2+xh+xh+h^2) +4x+4h+5 \\
&=3(x^2+2xh+h^2)+4x+4h+5 \\
&=3x^2+6xh+3h^2+4x+4h+5
\end{align}

Then $f(x+h)-f(x)$:

\begin{align}
f(x+h)-f(x)&=3x^2+6xh+3h^2+4x+4h+5-(3x^2+4x+5) \\
&=3x^2+6xh+3h^2+4x+4h+5-3x^2-4x-5 \\
&=6xh+3h^2+4h
\end{align}

#### Final Answer:

$$6xh+3h^2+4h$$

:::

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<summary> Solution: </summary>

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

First we find and simplify $f(x+h)$:

\begin{align}
f(x+h)&=2(x+h)^2+2(x+h)+4 \\
&=2(x^2+2xh+h^2)+2x+2h+4 \\
&=2x^2+4xh+2h^2+2x+2h+4
\end{align}

Then $f(x+h)-f(x)$:

\begin{align}
f(x+h)-f(x)&=2x^2+4xh+2h^2+2x+2h+4-(2x^2+2x+4) \\
&=2x^2+4xh+2h^2+2x+2h+4-2x^2-2x-4 \\
&=4xh+2h^2+2h
\end{align}

#### Final Answer:

$$4xh+2h^2+2h$$

:::

### b. $g(x)=-x^2+2x+5$ ; $g(x+h)-g(x)$

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<summary> Example: </summary>

Let $g(x) = -2x^2 + 3x + 4$. Now find $g(x+h) - g(x)$.

First, we find and simplify $g(x+h)$:

\begin{align}
g(x+h) &= -2(x+h)^2 + 3(x+h) + 4 \\
&= -2(x+h)(x+h) + 3(x+h) + 4 \\
&= -2(x^2 + 2xh + h^2) + 3x + 3h + 4 \\
&= -2x^2 - 4xh - 2h^2 + 3x + 3h + 4
\end{align}

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

\begin{align}
g(x+h) - g(x) &= \left(-2x^2 - 4xh - 2h^2 + 3x + 3h + 4\right) - \left(-2x^2 + 3x + 4\right) \\
&= -2x^2 - 4xh - 2h^2 + 3x + 3h + 4 + 2x^2 - 3x - 4 \\
&= -4xh - 2h^2 + 3h
\end{align}

#### Final Answer:

$$-4xh - 2h^2 + 3h$$

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<summary> Solution: </summary>

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

First, we find and simplify $g(x+h)$:

\begin{align}
g(x+h) &= -(x+h)^2+2(x+h)+5 \\
&=-(x^2+2xh+h^2)+2x+2h+5 \\
&=-x^2-2xh-h^2+2x+2h+5 \\
\end{align}

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

\begin{align}
g(x+h) - g(x) &= \left(-x^2-2xh-h^2+2x+2h+5\right) - \left(-x^2+2x+5\right) \\
&= -x^2-2xh-h^2+2x+2h+5+x^2-2x-5 \\
&= -2xh-h^2+2h
\end{align}

#### Final Answer:

$$-2xh-h^2+2h$$

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### c. $k(x)=6x^2-2x+3$ ; $k(x+h)-k(x)$

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<summary> Example: </summary>

Let $k(x) = 5x^2 - 4x + 2$. Now find $k(x+h) - k(x)$.

First, we find and simplify $k(x+h)$:

\begin{align}
k(x+h) &= 5(x+h)^2 - 4(x+h) + 2 \\
&= 5(x+h)(x+h) - 4(x+h) + 2 \\
&= 5(x^2 + 2xh + h^2) - 4x - 4h + 2 \\
&= 5x^2 + 10xh + 5h^2 - 4x - 4h + 2
\end{align}

Now, subtract $k(x)$ from $k(x+h)$:

\begin{align}
k(x+h) - k(x) &= \left(5x^2 + 10xh + 5h^2 - 4x - 4h + 2\right) - \left(5x^2 - 4x + 2\right) \\
&= 5x^2 + 10xh + 5h^2 - 4x - 4h + 2 - 5x^2 + 4x - 2 \\
&= 10xh + 5h^2 - 4h
\end{align}

#### Final Answer:

$$10xh + 5h^2 - 4h$$

:::

::: spoiler

<summary> Solution: </summary>

First, we find and simplify $k(x+h)$:

\begin{align}
k(x+h) &= 6(x+h)^2 - 2(x+h) + 3 \\
&= 6(x+h)(x+h) - 2(x+h) + 3 \\
&= 6(x^2 + 2xh + h^2) - 2(x+h) + 3 \\
&= 6x^2 + 12xh + 6h^2 - 2x - 2h + 3
\end{align}

Now, subtract $k(x)$ from $k(x+h)$:

\begin{align}
k(x+h) - k(x) &= \left(6x^2 + 12xh + 6h^2 - 2x - 2h + 3\right) - \left(6x^2 - 2x + 3\right) \\
&= 6x^2 + 12xh + 6h^2 - 2x - 2h + 3 - 6x^2 + 2x - 3 \\
&= 12xh + 6h^2 - 2h
\end{align}

#### Final Answer:

$12xh + 6h^2 - 2h$

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## Question 3

Find the indicated quantites for $f(x)=3x^2$:

### a. The slope of the secant line through the points $(1,f(1))$ and $(4,f(4))$ on the graph of $f(x)$.

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<summary> Example: </summary>

Find the indicated quantities for $f(x) = 2x^2$:

### a. The slope of the secant line through the points $(2,f(2))$ and $(3,f(3))$ on the graph of $f(x)$.

First, find the function values at $x = 2$ and $x = 3$:

\begin{align}
f(2) &= 2(2)^2 = 2 \cdot 4 = 8 \\
f(3) &= 2(3)^2 = 2 \cdot 9 = 18
\end{align}

Now, the slope of the secant line between the points $(2, f(2))$ and $(3, f(3))$ is given by the formula:

$$
\text{slope} = \frac{f(3) - f(2)}{3 - 2}
$$

Substitute the values we found:

\begin{align}
\text{slope} &= \frac{18 - 8}{3 - 2} \\
&= \frac{10}{1} = 10
\end{align}

#### Final Answer:

The slope of the secant line is $10$.

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<summary> Solution: </summary>

We are given $f(x) = 3x^2$. Now, we need to find the slope of the secant line through the points $(1, f(1))$ and $(4, f(4))$ on the graph of $f(x)$.

First, we find $f(1)$ and $f(4)$:

\begin{align}
f(1) &= 3(1)^2 = 3 \\
f(4) &= 3(4)^2 = 3(16) = 48
\end{align}

The slope of the secant line between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
$$

Substitute the values:

\begin{align}
\text{slope} &= \frac{f(4) - f(1)}{4 - 1} \\
&= \frac{48 - 3}{4 - 1} \\
&= \frac{45}{3} \\
&= 15
\end{align}

#### Final Answer:

The slope of the secant line is $15$.

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### b. The slope of the secant line through the points $(1,f(1))$ and $(1+h,f(1+h))$ on the graph of $f(x)$.

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<summary> Example: </summary>

Find the slope of the secant line through the points $(2, f(2))$ and $(2+h, f(2+h))$ on the graph of $f(x) = 2x^2$.

First, find $f(2)$:

$$
f(2) = 2(2)^2 = 2 \cdot 4 = 8
$$

Next, find $f(2+h)$:

\begin{align}
f(2+h) &= 2(2+h)^2 \\
&= 2(4 + 4h + h^2) \\
&= 2(4) + 2(4h) + 2(h^2) \\
&= 8 + 8h + 2h^2
\end{align}

Now, the slope of the secant line between the points $(2, f(2))$ and $(2+h, f(2+h))$ is given by the formula:

$$
\text{slope} = \frac{f(2+h) - f(2)}{(2+h) - 2}
$$

Substitute the values we found:

$$
\text{slope} = \frac{(8 + 8h + 2h^2) - 8}{h}
$$

Simplify the expression:

$$
\text{slope} = \frac{8h + 2h^2}{h}
$$

Finally, factor out $h$:

$$
\text{slope} = \dfrac{h(8+2h)}{h}= 8 + 2h
$$

#### Final Answer:

The slope of the secant line is $8 + 2h$.

:::

::: spoiler

<summary> Solution: </summary>

We are given $f(x) = 3x^2$. Now, we need to find the slope of the secant line through the points $(1, f(1))$ and $(1+h, f(1+h))$ on the graph of $f(x)$.

First, we find $f(1)$ and $f(1+h)$:

\begin{align}
f(1) &= 3(1)^2 = 3 \\
f(1+h) &= 3(1+h)^2 = 3(1 + 2h + h^2) = 3 + 6h + 3h^2
\end{align}

The slope of the secant line between the points $(1, f(1))$ and $(1+h, f(1+h))$ is given by:

$$
\text{slope} = \frac{f(1+h) - f(1)}{(1+h) - 1}
$$

Substitute the values:

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

#### Final Answer:

The slope of the secant line is $6 + 3h$.

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## Question 4

Use the four-step process to find $f'(x)$ and then find $f'(1)$ for each $f(x)$.

### a. $f(x)=2x^2+8$

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<summary> Example: </summary>

To find $f'(x)=\displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ using the definition of the derivative, where $f(x) = 3x^2 + 5$, we use the four-step process.

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

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

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

\begin{align}
f(x+h) - f(x) &= \left(3x^2 + 6xh + 3h^2 + 5\right) - \left(3x^2 + 5\right) \\
&= 3x^2 + 6xh + 3h^2 + 5 - 3x^2 - 5 \\
&= 6xh + 3h^2
\end{align}

### Step 3: Compute $\frac{f(x+h) - f(x)}{h}$

\begin{align}
\frac{f(x+h) - f(x)}{h} &= \frac{6xh + 3h^2}{h} \\
&= 6x + 3h
\end{align}

### Step 4: Take the limit as $h \to 0$

\begin{align}
f'(x) &= \lim_{h \to 0} (6x + 3h) \\
&= 6x
\end{align}

#### Final Answer:

$$ f'(x) = 6x $$

Then $f'(1)=6(1)=6$.

:::

::: spoiler

<summary> Solution: </summary>

We are given $f(x) = 2x^2 + 8$. We will use the four-step process to find $f'(x)=\displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ and then find $f'(1)$.

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

\begin{align}
f(x+h) &= 2(x+h)^2 + 8 \\
&= 2(x^2 + 2xh + h^2) + 8 \\
&= 2x^2 + 4xh + 2h^2 + 8
\end{align}

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

\begin{align}
f(x+h) - f(x) &= \left( 2x^2 + 4xh + 2h^2 + 8 \right) - \left( 2x^2 + 8 \right) \\
&= 4xh + 2h^2
\end{align}

### Step 3: Divide by $h$

\begin{align}
\frac{f(x+h) - f(x)}{h} &= \frac{4xh + 2h^2}{h} \\
&= 4x + 2h
\end{align}

### Step 4: Take the limit as $h$ approaches 0

\begin{align}
f'(x) &= \lim_{h \to 0} (4x + 2h) \\
&= 4x
\end{align}

Thus, $f'(x) = 4x$.

Now, find $f'(1)$:

\begin{align}
f'(1) &= 4(1) = 4
\end{align}

#### Final Answer:

$f'(x) = 4x$ and $f'(1) = 4$

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### b. $f(x)=\dfrac{6}{x}-2$

::: spoiler

<summary> Example: </summary>

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

To find $f'(x)$ using the definition of the derivative, where $f(x) = \dfrac{3}{x} - 1$, we use the four-step process.

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

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

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

\begin{align}
f(x+h) - f(x) &= \left(\frac{3}{x+h} - 1\right) - \left(\frac{3}{x} - 1\right) \\
&= \frac{3}{x+h} - 1 - \frac{3}{x} + 1 \\
&= \frac{3}{x+h} - \frac{3}{x} \\
&= \frac{3}{x+h} \left(\dfrac{x}{x}\right) - \frac{3}{x}\left(\dfrac{x+h}{x+h}\right) \\
&= \frac{3x - 3(x+h)}{x(x+h)} \\
&= \frac{3x - 3x - 3h}{x(x+h)} \\
&= \frac{-3h}{x(x+h)}
\end{align}

### Step 3: Compute $\frac{f(x+h) - f(x)}{h}$

\begin{align}
\frac{f(x+h) - f(x)}{h} &= \frac{-3h}{h \cdot x(x+h)} \\
&= \frac{-3}{x(x+h)}
\end{align}

### Step 4: Take the limit as $h \to 0$

\begin{align}
f'(x) &= \lim_{h \to 0} \frac{-3}{x(x+h)} \\
&=\dfrac{-3}{x(x+0)} \\
&= \frac{-3}{x^2}
\end{align}

#### Final Answer:

$$ f'(x) = \frac{-3}{x^2} $$

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::: spoiler

<summary> Solution: </summary>

We are given $f(x) = \dfrac{6}{x} - 2$. We will use the four-step process to find $f'(x)$.

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

\begin{align}
f(x+h) &= \dfrac{6}{x+h} - 2
\end{align}

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

\begin{align}
f(x+h) - f(x) &= \left( \dfrac{6}{x+h} - 2 \right) - \left( \dfrac{6}{x} - 2 \right) \\
&= \dfrac{6}{x+h} - \dfrac{6}{x} \\
&= \dfrac{6}{x+h}\left(\dfrac{x}{x}\right) - \dfrac{6}{x}\left(\dfrac{x+h}{x+h}\right) \\
\end{align}

We now simplify the difference between the two fractions. Find a common denominator:

\begin{align}
f(x+h) - f(x) &= \dfrac{6x - 6(x+h)}{x(x+h)} \\
&= \dfrac{6x - 6x - 6h}{x(x+h)} \\
&= \dfrac{-6h}{x(x+h)}
\end{align}

### Step 3: Divide by $h$

\begin{align}
\frac{f(x+h) - f(x)}{h} &= \frac{\dfrac{-6h}{x(x+h)}}{h} \\
&= \dfrac{-6}{x(x+h)}
\end{align}

### Step 4: Take the limit as $h$ approaches $0$

\begin{align}
f'(x) &= \lim_{h \to 0} \dfrac{-6}{x(x+h)} \\
&= \dfrac{-6}{x^2}
\end{align}

Thus, $f'(x) = \dfrac{-6}{x^2}$.

#### Final Answer:

$f'(x) = \dfrac{-6}{x^2}$

$f'(1)=\dfrac{-6}{1^2}=-6$

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