# Math 105 Exam 2 Practice

## Formulas:

$$\frac{y_2-y_1}{x_2-x_1}$$

$$y-y_1=m(x-x_1)$$

$$P(x)=R(x)-C(x)$$

---

## Question 1. Graph the equation using the slope and $y$-intercept. $y=-\dfrac{1}{2}x-2$


<details> <summary> Solution: </summary>

The equation of the line is in $y=mx+b$ form, where $m$ is the slope of the line, and $b$ is the $y$-coordinate of the $y$-intercept.

In this case we have $b=-2$ and $m=-\dfrac{1}{2}$.

The $y$-intercept is $(0,-2)$ and the slope is $\dfrac{-1}{2}$, so right 2 and down 1.


</details>

## Question 2. Find the $y=mx+b$ form for the equation of the line through $(3,-7)$ with slope $-\dfrac{1}{4}$.

<details> <summary> Solution: </summary>

We are given the point $(3, -7)$ and the slope $m = -\dfrac{1}{4}$. To find the equation of the line in $y = mx + b$ form, we use the point-slope formula:

$$
y - y_1 = m(x - x_1)
$$

Substituting the given values $(x_1, y_1) = (3, -7)$ and $m = -\dfrac{1}{4}$:

$$
y - (-7) = -\dfrac{1}{4}(x - 3)
$$

This simplifies to:

$$
y + 7 = -\dfrac{1}{4}(x - 3)
$$

Distribute the slope:

$$
y + 7 = -\dfrac{1}{4}x + \dfrac{3}{4}
$$

Now, subtract $7$ from both sides:

$$
y = -\dfrac{1}{4}x + \dfrac{3}{4} - 7
$$

Convert $7$ to a fraction with denominator 4:

$$
y = -\dfrac{1}{4}x + \dfrac{3}{4} - \dfrac{28}{4}
$$

Simplify:

$$
y = -\dfrac{1}{4}x - \dfrac{25}{4}
$$

Thus, the equation of the line is:

$$
y = -\dfrac{1}{4}x - \dfrac{25}{4}
$$

</details>

## Question 3. Consider the line $2x-7y=8$.

### a. What is its slope?

<details> <summary> Solution: </summary>

We are given the equation of the line in standard form:

$$
2x - 7y = 8
$$

To find the slope, we need to rewrite the equation in slope-intercept form, $y = mx + b$, where $m$ is the slope.

Starting with:

$$
2x - 7y = 8
$$

Solve for $y$ by isolating it on one side:

$$
-7y = -2x + 8
$$

Now, divide both sides by $-7$:

$$
y = \dfrac{2}{7}x - \dfrac{8}{7}
$$

Thus, the equation of the line in slope-intercept form is:

$$
y = \dfrac{2}{7}x - \dfrac{8}{7}
$$

From this, we can see that the slope of the line is:

$$
m = \dfrac{2}{7}
$$

</details>

### b. What is the equation of the line perpendicular to this line through $(1,-5)$?

<details> <summary> Solution: </summary>

We are given the line with equation:

$$
2x - 7y = 8
$$

First, we rewrite this in slope-intercept form to find its slope:

$$
2x - 7y = 8
$$

Solve for $y$:

$$
-7y = -2x + 8
$$

Divide by $-7$:

$$
y = \dfrac{2}{7}x - \dfrac{8}{7}
$$

The slope of this line is $m = \dfrac{2}{7}$.

### Finding the slope of the perpendicular line

The slope of a line perpendicular to another is the negative reciprocal of the original slope. So, the slope of the perpendicular line is:

$$
m_{\text{perpendicular}} = -\dfrac{7}{2}
$$

### Finding the equation of the perpendicular line through the point $(1, -5)$

We use the point-slope form of a line equation:

$$
y - y_1 = m(x - x_1)
$$

Substituting $m = -\dfrac{7}{2}$ and the point $(1, -5)$:

$$
y - (-5) = -\dfrac{7}{2}(x - 1)
$$

Simplify:

$$
y + 5 = -\dfrac{7}{2}(x - 1)
$$

Distribute the slope:

$$
y + 5 = -\dfrac{7}{2}x + \dfrac{7}{2}
$$

Now, subtract $5$ from both sides:

$$
y = -\dfrac{7}{2}x + \dfrac{7}{2} - 5
$$

Convert $5$ to a fraction with denominator 2:

$$
y = -\dfrac{7}{2}x + \dfrac{7}{2} - \dfrac{10}{2}
$$

Simplify:

$$
y = -\dfrac{7}{2}x - \dfrac{3}{2}
$$

Thus, the equation of the line perpendicular to $2x - 7y = 8$ through the point $(1, -5)$ is:

$$
y = -\dfrac{7}{2}x - \dfrac{3}{2}
$$

</details>

## Question 4. Consider the line $2x-9y=8$

### a. What is the slope of this line?

<details> <summary> Solution: </summary>

We are given the equation of the line in standard form:

$$
2x - 9y = 8
$$

To find the slope, we need to rewrite the equation in slope-intercept form, $y = mx + b$, where $m$ is the slope.

Starting with:

$$
2x - 9y = 8
$$

Solve for $y$ by isolating it on one side:

$$
-9y = -2x + 8
$$

Now, divide both sides by $-9$:

$$
y = \dfrac{2}{9}x - \dfrac{8}{9}
$$

Thus, the equation of the line in slope-intercept form is:

$$
y = \dfrac{2}{9}x - \dfrac{8}{9}
$$

From this, we can see that the slope of the line is:

$$
m = \dfrac{2}{9}
$$

</details>

### b. What is the equation of the line parallel to this line through $(1,-5)$?

<details> <summary> Solution: </summary>

We are given the line with equation:

$$
2x - 9y = 8
$$

First, we rewrite this in slope-intercept form to find its slope:

$$
2x - 9y = 8
$$

Solve for $y$:

$$
-9y = -2x + 8
$$

Divide by $-9$:

$$
y = \dfrac{2}{9}x - \dfrac{8}{9}
$$

The slope of this line is $m = \dfrac{2}{9}$.

### Finding the equation of the parallel line through the point $(1, -5)$

The slope of a parallel line is the same as the slope of the given line. So, the slope of the parallel line is:

$$
m_{\text{parallel}} = \dfrac{2}{9}
$$

Now, using the point-slope form of a line equation:

$$
y - y_1 = m(x - x_1)
$$

Substituting $m = \dfrac{2}{9}$ and the point $(1, -5)$:

$$
y - (-5) = \dfrac{2}{9}(x - 1)
$$

Simplify:

$$
y + 5 = \dfrac{2}{9}(x - 1)
$$

Distribute the slope:

$$
y + 5 = \dfrac{2}{9}x - \dfrac{2}{9}
$$

Now, subtract $5$ from both sides:

$$
y = \dfrac{2}{9}x - \dfrac{2}{9} - 5
$$

Convert $5$ to a fraction with denominator 9:

$$
y = \dfrac{2}{9}x - \dfrac{2}{9} - \dfrac{45}{9}
$$

Simplify:

$$
y = \dfrac{2}{9}x - \dfrac{47}{9}
$$

Thus, the equation of the line parallel to $2x - 9y = 8$ through the point $(1, -5)$ is:

$$
y = \dfrac{2}{9}x - \dfrac{47}{9}
$$

</details>

## Question 5. Consider the points $(2,4)$ and $(-8,7)$.

### a. What is the slope through these two points?

<details> <summary> Solution: </summary>

We are given the points $(2, 4)$ and $(-8, 7)$.

The formula for the slope $m$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$
m = \dfrac{y_2 - y_1}{x_2 - x_1}
$$

Substitute the given points $(x_1, y_1) = (2, 4)$ and $(x_2, y_2) = (-8, 7)$:

$$
m = \dfrac{7 - 4}{-8 - 2} = \dfrac{3}{-10} = -\dfrac{3}{10}
$$

So, the slope of the line is:

$$
m = -\dfrac{3}{10}
$$

</details>

### b. What is the equation of the line through these points?

<details> <summary> Solution: </summary>

We are given the points $(2, 4)$ and $(-8, 7)$ and the slope $m = -\dfrac{3}{10}$ from part a. Now, we need to find the equation of the line passing through these points.

### Step 1: Use the point-slope form

The point-slope form of a line is:

$$
y - y_1 = m(x - x_1)
$$

We can use either point. Let's use the point $(2, 4)$ and the slope $m = -\dfrac{3}{10}$.

Substitute these values into the point-slope formula:

$$
y - 4 = -\dfrac{3}{10}(x - 2)
$$

### Step 2: Simplify to slope-intercept form

Distribute the slope:

$$
y - 4 = -\dfrac{3}{10}x + \dfrac{6}{10}
$$

Simplify the constant term:

$$
y - 4 = -\dfrac{3}{10}x + \dfrac{3}{5}
$$

Now, add 4 to both sides:

$$
y = -\dfrac{3}{10}x + \dfrac{3}{5} + 4
$$

Convert 4 to a fraction with denominator 5:

$$
y = -\dfrac{3}{10}x + \dfrac{3}{5} + \dfrac{20}{5}
$$

Simplify:

$$
y = -\dfrac{3}{10}x + \dfrac{23}{5}
$$

Thus, the equation of the line passing through the points $(2, 4)$ and $(-8, 7)$ is:

$$
y = -\dfrac{3}{10}x + \dfrac{23}{5}
$$

</details>

## Question 6. Find the equation of the line through $(4,-2)$ parallel to the line $x=-4$. (Show work in the graph.)


<details> <summary> Solution: </summary>

We are given the point $(4, -2)$ and asked to find the equation of a line that is parallel to the line $x = -4$.

### Step 1: Understand the line $x = -4$

The line $x = -4$ is a vertical line passing through $x = -4$. Any line parallel to this will also be a vertical line. Vertical lines have the form:

$$
x = c
$$

where $c$ is a constant.

### Step 2: Equation of the parallel line

Since the line must be parallel to $x = -4$, it will also be a vertical line, but it will pass through the point $(4, -2)$. The equation of a vertical line passing through $x = 4$ is:

$$
x = 4
$$

Thus, the equation of the line parallel to $x = -4$ passing through the point $(4, -2)$ is:

$$
x = 4
$$


</details>

## Question 7. Find the equation of the line through $(4,-2)$ perpendicular to the line $x=-4$. (Show work in the graph.)


<details> <summary> Solution: </summary>

We are given the point $(4, -2)$ and asked to find the equation of a line that is perpendicular to the line $x = -4$.

### Step 1: Understand the line $x = -4$

The line $x = -4$ is a vertical line. A line that is perpendicular to a vertical line will be a horizontal line. Horizontal lines have the form:

$$
y = c
$$

where $c$ is a constant.

### Step 2: Equation of the perpendicular line

Since the line must be perpendicular to $x = -4$, it will be a horizontal line passing through the point $(4, -2)$. The equation of a horizontal line passing through $y = -2$ is:

$$
y = -2
$$

Thus, the equation of the line perpendicular to $x = -4$ passing through the point $(4, -2)$ is:

$$
y = -2
$$


</details>

## Question 8. Solve the equation $2(2x+1)-3(-x+5)=4(3x-1)$

<details> <summary> Solution: </summary>

| **Equation** | **Reason/Explanation** |
|--------------------------------------------|----------------------------------------------------------|
| $2(2x + 1) - 3(-x + 5) = 4(3x - 1)$ | Start with the given equation. |
| $2 \cdot 2x + 2 \cdot 1 - 3 \cdot (-x) - 3 \cdot 5 = 4 \cdot 3x - 4 \cdot 1$ | Apply distribution to each term inside parentheses. |
| $4x + 2 + 3x - 15 = 12x - 4$ | Simplify each term after distributing. |
| $7x - 13 = 12x - 4$ | Combine like terms on both sides. |
| $7x - 12x - 13 = -4$ | Subtract $12x$ from both sides. |
| $-5x - 13 = -4$ | Simplify the left side. |
| $-5x - 13 + 13 = -4 + 13$ | Add 13 to both sides to isolate $x$ terms. |
| $-5x = 9$ | Simplify both sides. |
| $x = -\dfrac{9}{5}$ | Solve for $x$ by dividing both sides by $-5$. |

Thus, the solution is $x=-\dfrac{9}{5}$.

</details>

## Question 9. Solve the equation. $3(2x+1)-2(x+5)=4(x-3)$

<details> <summary> Solution: </summary>

| **Equation** | **Reason/Explanation** |
|--------------------------------------------|----------------------------------------------------------|
| $3(2x + 1) - 2(x + 5) = 4(x - 3)$ | Start with the given equation. |
| $3 \cdot 2x + 3 \cdot 1 - 2 \cdot x - 2 \cdot 5 = 4 \cdot x - 4 \cdot 3$ | Apply distribution to each term inside parentheses. |
| $6x + 3 - 2x - 10 = 4x - 12$ | Simplify each term after distributing. |
| $4x - 7 = 4x - 12$ | Combine like terms on both sides. |
| $4x - 4x - 7 = -12$ | Subtract $4x$ from both sides. |
| $-7 = -12$ | Simplify the left side. |
| $-7 \neq -12$ | Contradiction—no solution exists. |

</details>

## Question 10. Solve the inequality. Write the solution in interval notation. $1-2(3x-1) \leq 2x+5$

<details> <summary> Solution </summary>

| **Equation** | **Reason/Explanation** |
|--------------------------------------------|----------------------------------------------------------|
| $1 - 2(3x - 1) \leq 2x + 5$ | Start with the given inequality. |
| $1 - 2 \cdot 3x + 2 \cdot 1 \leq 2x + 5$ | Apply distribution to the term inside parentheses. |
| $1 - 6x + 2 \leq 2x + 5$ | Simplify the distribution. |
| $3 - 6x \leq 2x + 5$ | Combine like terms on the left side. |
| $3 - 5 \leq 2x + 6x$ | Subtract $2x$ from both sides to move $x$ terms to the left side. |
| $3 - 6x - 2x \leq 5$ | Simplify the terms involving $x$. |
| $3 - 8x \leq 5$ | Combine the $x$ terms on the left side. |
| $-8x \leq 5 - 3$ | Subtract $3$ from both sides to move constants to the right. |
| $-8x \leq 2$ | Simplify the constants on the right side. |
| $x \geq \dfrac{2}{-8}$ | Divide both sides by $-8$, reversing the inequality sign because we divided by a negative number. |
| $x \geq -\dfrac{1}{4}$ | Simplify the fraction. |

The solution to the inequality is the interval $$\left[ -\dfrac{1}{4},\infty\right)$$.

</details>

## Question 11. Find the domain of $f(x)=\sqrt{5x+9}$.

<details> <summary> Solution: </summary>

| **Step** | **Explanation** |
|--------------------------------------------|----------------------------------------------------------|
| $f(x) = \sqrt{5x + 9}$ | Start with the given function. |
| $5x + 9 \geq 0$ | The expression inside the square root must be non-negative (since the square root of a negative number is undefined for real numbers). |
| $5x \geq -9$ | Subtract 9 from both sides. |
| $x \geq -\dfrac{9}{5}$ | Divide both sides by 5 to solve for $x$. |
| Domain: $x \geq -\dfrac{9}{5}$ | The domain is all values of $x$ such that $x \geq -\dfrac{9}{5}$. |

Thus, the domain of the function $f(x) = \sqrt{5x + 9}$ is the interval $\left[-\dfrac{9}{5},\infty\right)$.

</details>

## Question 12. Find the domain of $f(x)=\dfrac{2}{\sqrt{3x+11}}$

<details> <summary> Solution: </summary>

| **Step** | **Explanation** |
|--------------------------------------------|----------------------------------------------------------|
| $f(x) = \dfrac{2}{\sqrt{3x + 11}}$ | Start with the given function. |
| $3x + 11 > 0$ | The expression inside the square root must be positive because the denominator cannot be zero, and the square root of a negative number is undefined. |
| $3x > -11$ | Subtract 11 from both sides. |
| $x > -\dfrac{11}{3}$ | Divide both sides by 3 to solve for $x$. |
| Domain: $x > -\dfrac{11}{3}$ | The domain is all values of $x$ such that $x > -\dfrac{11}{3}$. |

Thus, the domain of the function $f(x) = \dfrac{2}{\sqrt{3x + 11}}$ is the interval $\left(-\dfrac{11}{3},\infty\right)$.

</details>

## Question 13. A graph of a function $f(x)$ is shown in Figure 1. Using the graph, state the intervals where $f(x)$ is increasing and decreasing. Also state the relative maximum and minimum values for $f(x)$.

Increasing:
Decreasing:
Relative Maximum:
Relative Minimum:


<details> <summary> Solution: </summary>

Based on the graph of the function $f(x)$, we can determine the following:

### Intervals where $f(x)$ is increasing:
- The function is increasing on the interval:
- From $x = -2$ to $x = 0$
- From $x = 1$ to $x = \infty$

- Thus $f(x)$ is increasing on intervals $(-2,0) \cup (1,\infty)$.

### Intervals where $f(x)$ is decreasing:
- The function is decreasing on the intervals:
- From $x = -\infty$ to $x = -2$
- From $x = 0$ to $x = 1$

- Thus $f(x)$ is decreasing on intervals $(-\infty,-2) \cup (0,1)$.

### Relative maximum:
- There is a relative maximum at $x = 0$, where the function value is $y = 3$.

### Relative minimum:
- There is a relative minimums at $x = -2$, where the function value is $y = 1$.
- There is another relative minimum at $x=1$, where the function value is $y=-1$.

</details>

## Question 14. A graph of a function $f(x)$ is shown in Figure 1. Using the graph, state the intervals where $f(x)$ is increasing, decreasing, and constant.


<details> <summary> Solution: </summary>

Based on the graph of the function $f(x)$, we can determine the following:

### Intervals where $f(x)$ is increasing:
- The function is increasing on the interval:
- From $x = 1$ to $x = 2$

- Thus $f(x)$ is increasing on interval $(1,2)$.

### Intervals where $f(x)$ is decreasing:
- The function is decreasing on the interval:
- From $x = -2$ to $x = 0$

- Thus $f(x)$ is decreasing on interval $(-2,0)$.

### Intervals where $f(x)$ is constant:
- The function is constant on the intervals:
- From $x = -4$ to $x = -2$
- From $x = 0$ to $x = 1$

- Thus $f(x)$ is constant on intervals $(-4,-2) \cup (0,1)$.

</details>

## Question 15. Let $$f(x)=\begin{cases} \dfrac{1}{3}x-2 & \text{ if } x \leq 0 \\ x^2 & \text{ if } 0<x<3 \\ 2x+3 & \text{ if } x \geq 3 \end{cases}$$

### a. Find $f(5)$

<details> <summary> Solution: </summary>

We are asked to find $f(5)$. Let's evaluate the inequalities for each case and determine which condition is true.

| **Piece** | **Inequality** | **Substitute $x = 5$** | **True/False** |
|-------------------------------------|-----------------------------------|-------------------------------|-------------------------------|
| $\dfrac{1}{3}x - 2$ | $x \leq 0$ | $5 \leq 0$ | False |
| $x^2$ | $0 < x < 3$ | $0 < 5 < 3$ | False |
| $2x + 3$ | $x \geq 3$ | $5 \geq 3$ | True |

### Conclusion:
Since the inequality $x \geq 3$ is true for $x = 5$, we use the third case of the piecewise function, $f(x) = 2x + 3$.

Now, substitute $x = 5$ into this expression:

$$
f(5) = 2(5) + 3 = 10 + 3 = 13
$$

Thus, $f(5) = 13$.

</details>

### b. Find $f(2)$

<details> <summary> Solution: </summary>

We are asked to find $f(2)$. Let's evaluate the inequalities for each case and determine which condition is true.

| **Piece** | **Inequality** | **Substitute $x = 2$** | **True/False** |
|-------------------------------------|-----------------------------------|-------------------------------|-------------------------------|
| $\dfrac{1}{3}x - 2$ | $x \leq 0$ | $2 \leq 0$ | False |
| $x^2$ | $0 < x < 3$ | $0 < 2 < 3$ | True |
| $2x + 3$ | $x \geq 3$ | $2 \geq 3$ | False |

### Conclusion:
Since the inequality $0 < x < 3$ is true for $x = 2$, we use the second case of the piecewise function, $f(x) = x^2$.

Now, substitute $x = 2$ into this expression:

$$
f(2) = 2^2 = 4
$$

Thus, $f(2) = 4$.

</details>

### c. Find $f(-4)$

<details> <summary> Solution: </summary>

We are asked to find $f(-4)$. Let's evaluate the inequalities for each case and determine which condition is true.

| **Piece** | **Inequality** | **Substitute $x = -4$** | **True/False** |
|-------------------------------------|-----------------------------------|-------------------------------|-------------------------------|
| $\dfrac{1}{3}x - 2$ | $x \leq 0$ | $-4 \leq 0$ | True |
| $x^2$ | $0 < x < 3$ | $0 < -4 < 3$ | False |
| $2x + 3$ | $x \geq 3$ | $-4 \geq 3$ | False |

### Conclusion:
Since the inequality $x \leq 0$ is true for $x = -4$, we use the first case of the piecewise function, $f(x) = \dfrac{1}{3}x - 2$.

Now, substitute $x = -4$ into this expression:

$$
f(-4) = \dfrac{1}{3}(-4) - 2 = -\dfrac{4}{3} - 2 = -\dfrac{4}{3} - \dfrac{6}{3} = -\dfrac{10}{3}
$$

Thus, $f(-4) = -\dfrac{10}{3}$.

</details>

## Question 16. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate $(f+g)(2)$ and $(fg)(-1)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(f+g)(2)$

1. We are asked to find $(f+g)(2)$.
2. This means we need to plug $x = 2$ into both $f(x)$ and $g(x)$ separately, then add the results.

- First, find $f(2)$:
$f(x) = x^2 + 2$

Plug $x = 2$:

$f(2) = 2^2 + 2 = 4 + 2 = 6$

- Next, find $g(2)$:
$g(x) = 3x - 2$

Plug $x = 2$:

$g(2) = 3(2) - 2 = 6 - 2 = 4$

3. Now, add the two results:
$f(2) + g(2) = 6 + 4 = 10$

So, $(f+g)(2) = 10$.

### Part 2: Calculating $(fg)(-1)$

1. We are asked to find $(fg)(-1)$.
2. This means we need to plug $x = -1$ into both $f(x)$ and $g(x)$, then multiply the results.

- First, find $f(-1)$:
$f(x) = x^2 + 2$

Plug $x = -1$:

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

- Next, find $g(-1)$:
$g(x) = 3x - 2$

Plug $x = -1$:

$g(-1) = 3(-1) - 2 = -3 - 2 = -5$

3. Now, multiply the two results:
$f(-1) \cdot g(-1) = 3 \cdot (-5) = -15$

So, $(fg)(-1) = -15$.

</details>

## Question 17. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate $(f-g)(1)$ and $(f/g)(2)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(f-g)(1)$

1. We are asked to find $(f-g)(1)$.
2. This means we need to plug $x = 1$ into both $f(x)$ and $g(x)$ separately, then subtract the results.

- First, find $f(1)$:
$f(x) = x^2 + 2$

Plug $x = 1$:
$f(1) = 1^2 + 2 = 1 + 2 = 3$

- Next, find $g(1)$:
$g(x) = 3x - 2$

Plug $x = 1$:
$g(1) = 3(1) - 2 = 3 - 2 = 1$

3. Now, subtract the two results:
$f(1) - g(1) = 3 - 1 = 2$

So, $(f-g)(1) = 2$.

### Part 2: Calculating $(f/g)(2)$

1. We are asked to find $(f/g)(2)$.
2. This means we need to plug $x = 2$ into both $f(x)$ and $g(x)$ separately, then divide the results.

- First, find $f(2)$:
$f(x) = x^2 + 2$

Plug $x = 2$:
$f(2) = 2^2 + 2 = 4 + 2 = 6$

- Next, find $g(2)$:
$g(x) = 3x - 2$

Plug $x = 2$:
$g(2) = 3(2) - 2 = 6 - 2 = 4$

3. Now, divide the two results:
$\dfrac{f(2)}{g(2)} = \dfrac{6}{4} = \dfrac{3}{2}$

So, $(f/g)(2) = \dfrac{3}{2}$.

</details>

## Question 18. Let $f(x)=x-5$ and $g(x)=3x+7$. Determine and simplify $(f+g)(x)$ and $(fg)(x)$.

<details> <summary> Solution: </summary>

We are given two functions:

$f(x) = x - 5$
$g(x) = 3x + 7$

### Part 1: Calculating $(f+g)(x)$

1. We are asked to find $(f+g)(x)$.
2. This means we need to add $f(x)$ and $g(x)$ together.

- First, write the sum of the two functions:
$(f+g)(x) = f(x) + g(x)$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(f+g)(x) = (x - 5) + (3x + 7)$

- Combine like terms:
$= x + 3x - 5 + 7$
$= 4x + 2$

So, $(f+g)(x) = 4x + 2$.

### Part 2: Calculating $(fg)(x)$ Using FOIL

1. We are asked to find $(fg)(x)$.
2. This means we need to multiply $f(x)$ and $g(x)$ together using FOIL (First, Outer, Inner, Last).

- Write the product:
$(fg)(x) = (x - 5)(3x + 7)$

Now, apply FOIL:

- **First** terms: $x \cdot 3x = 3x^2$
- **Outer** terms: $x \cdot 7 = 7x$
- **Inner** terms: $-5 \cdot 3x = -15x$
- **Last** terms: $-5 \cdot 7 = -35$

Now, combine all the terms:

$3x^2 + 7x - 15x - 35$

Combine like terms:

$= 3x^2 - 8x - 35$

So, $(fg)(x) = 3x^2 - 8x - 35$.

</details>

## Question 19. Let $f(x)=x^3$ and $g(x)=2x-5$. Determine and simplify $(f-g)(x)$ and $(fg)(x)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(f-g)(x)$

1. We are asked to find $(f-g)(x)$.
2. This means we need to subtract $g(x)$ from $f(x)$.

- First, write the difference of the two functions:
$(f-g)(x) = f(x) - g(x)$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(f-g)(x) = (x^3) - (2x - 5)$

- Simplify (distribute the negative):
$(f-g)(x) = x^3 - 2x + 5$

So, $(f-g)(x) = x^3 - 2x + 5$.

### Part 2: Calculating $(fg)(x)$.

1. We are asked to find $(fg)(x)$.
2. This means we need to multiply $f(x)$ and $g(x)$ together.

- Write the product:
$(fg)(x) = (x^3)(2x - 5)$

Now, distribute $x^3$ across the terms in $g(x)$:

- $x^3 \cdot 2x = 2x^4$
- $x^3 \cdot (-5) = -5x^3$

Thus, $(fg)(x) = 2x^4 - 5x^3$.

</details>

## Question 20. Let $f(x)=x^4$ and $g(x)=2x^5-3x^2+4$. Determine and simplify $(fg)(x)$ and $(g/f)(x)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(fg)(x)$

1. We are asked to find $(fg)(x)$.
2. This means we need to multiply $f(x)$ and $g(x)$ together.

- Write the product:
$(fg)(x) = f(x) \cdot g(x)$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(fg)(x) = (x^4)(2x^5 - 3x^2 + 4)$

Now, distribute $x^4$ across the terms in $g(x)$:

- $x^4 \cdot 2x^5 = 2x^9$
- $x^4 \cdot (-3x^2) = -3x^6$
- $x^4 \cdot 4 = 4x^4$

So, $(fg)(x) = 2x^9 - 3x^6 + 4x^4$.

### Part 2: Calculating $(g/f)(x)$

1. We are asked to find $(g/f)(x)$.
2. This means we need to divide $g(x)$ by $f(x)$.

- Write the quotient:
$(g/f)(x) = \dfrac{g(x)}{f(x)}$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(g/f)(x) = \dfrac{2x^5 - 3x^2 + 4}{x^4}$

Now, simplify each term by dividing by $x^4$:

- $\dfrac{2x^5}{x^4} = 2x$
- $\dfrac{-3x^2}{x^4} = -3x^{-2}$
- $\dfrac{4}{x^4} = 4x^{-4}$

So, $(g/f)(x) = 2x-3x^{-2} + 4x^{-4}$.

</details>

## Question 21. Let $f(x)=x^3$ and $g(x)=3x^4-2x^3+2$. Determine and simplify $(ff)(x)$ and $(g/f)(x)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(ff)(x)$

1. We are asked to find $(ff)(x)$.
2. This means we need to multiply $f(x)$ by itself.

- Write the product:
$(ff)(x) = f(x) \cdot f(x)$

- Now, substitute the expression for $f(x)$:
$(ff)(x) = (x^3) \cdot (x^3)$

Now, multiply the terms:

- $x^3 \cdot x^3 = x^{6}$

So, $(ff)(x) = x^{6}$.

### Part 2: Calculating $(g/f)(x)$

1. We are asked to find $(g/f)(x)$.
2. This means we need to divide $g(x)$ by $f(x)$.

- Write the quotient:
$(g/f)(x) = \dfrac{g(x)}{f(x)}$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(g/f)(x) = \dfrac{3x^4 - 2x^3 + 2}{x^3}$

Now, simplify each term by dividing by $x^3$:

- $\dfrac{3x^4}{x^3} = 3x$
- $\dfrac{-2x^3}{x^3} = -2$
- $\dfrac{2}{x^3} = \dfrac{2}{x^3}$

So, $(g/f)(x) = 3x - 2 + 2x^{-3}$.

</details>

## Question 22. Let $f(x)=x^5$ and $g(x)=3x^4-2x^3+2$. Determine and simplify $(f-g)(x)$ and $(f/g)(x)$.

<details> <summary> Solution: </summary>

We are given two functions:

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

### Part 1: Calculating $(f-g)(x)$

1. We are asked to find $(f-g)(x)$.
2. This means we need to subtract $g(x)$ from $f(x)$.

- Write the difference:
$(f-g)(x) = f(x) - g(x)$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(f-g)(x) = (x^5) - (3x^4 - 2x^3 + 2)$

Now, distribute the negative sign across $g(x)$:

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

So, $(f-g)(x) = x^5 - 3x^4 + 2x^3 - 2$.

### Part 2: Calculating $(f/g)(x)$

1. We are asked to find $(f/g)(x)$.
2. This means we need to divide $f(x)$ by $g(x)$.

- Write the quotient:
$(f/g)(x) = \dfrac{f(x)}{g(x)}$

- Now, substitute the expressions for $f(x)$ and $g(x)$:
$(f/g)(x) = \dfrac{x^5}{3x^4 - 2x^3 + 2}$

We cannot simplify the division further, so this is the final result:

So, $(f/g)(x) = \dfrac{x^5}{3x^4 - 2x^3 + 2}$.

</details>

## Question 23. If the revenue from producing $x$ units is $R(x)=20x-2x^2$ and the cost from producing $x$ units is $C(x)=5x+2$, what is the profit $P(x)$?

<details> <summary> Solution: </summary>

We are given the following functions:

- Revenue: $R(x) = 20x - 2x^2$
- Cost: $C(x) = 5x + 2$

### Calculating the Profit Function $P(x)$

1. The profit $P(x)$ is the difference between revenue $R(x)$ and cost $C(x)$.

- Write the profit equation:
$P(x) = R(x) - C(x)$

- Now, substitute the expressions for $R(x)$ and $C(x)$:
$P(x) = (20x - 2x^2) - (5x + 2)$

2. Simplify by distributing the negative sign across $C(x)$:

- $P(x) = 20x - 2x^2 - 5x - 2$

3. Combine like terms:

- $P(x) = (20x - 5x) - 2x^2 - 2$
- $P(x) = 15x - 2x^2 - 2$

So, the profit function is:
$P(x) = 15x - 2x^2 - 2$.

Rewriting with descending powers of $x$:

$P(x)=-2x^2+15x-2$

</details>