# Math 105 Exam 2 Practice

## Exam Study Ritual:

**For each problem:**
1. Write and speak each problem and solution out loud.
2. Come up with notes to write to your future self to help you remember how to do that particular problem.
3. Be able to write and generate your own correct solution from scratch without looking at the solutions.
4. Make flash cards where the front is the question and the back is the solution. Look at every flash card multiple times every day.

**In general:**
1. _Act_ as if you are allowed one index card of notes on the exam. Create multiple revisions of the note card.
2. Get plenty of sleep.
3. Go to the restroom just before the exam or whatever you need to do so it's not an issue at all during the exam.

## 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. [PEMDAS](https://hackmd.io/@m105collegealgebra/review). Calculate

### a. $$-5 \cdot 2^3+2 (-2)^2$$

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

| Equation | Explanation |
|---------------------------------------------|-------------------------------------------------------------------|
| $$ -5 \cdot 2^3 + 2(-2)^2$$ | Original expression. |
| $$ -5 \cdot 8 + 2(-2)^2$$ | Replace $2^3$ with $8$ because $2^3 =2 \cdot 2 \cdot 2= 8$. |
| $$ -5 \cdot 8 + 2 \cdot 4$$ | Replace $(-2)^2$ with $4$ because $(-2)^2 =(-2)(-2)= 4$. |
| $$ -40 + 2 \cdot 4$$ | Compute $-5 \cdot 8$ to get $-40$. |
| $$ -40 + 8$$ | Compute $2 \cdot 4$ to get $8$. |
| $$ -32$$ | Add $-40$ and $8$ to obtain the final result $-32$. |

</details>

### b. $$\left(\dfrac{-4}{5}\right)\left(\dfrac{-3}{2}\right)$$

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

| Equation | Explanation |
|--------------------------------------------------------------------------|----------------------------------------------------------------------------------|
| $$\left(\frac{-4}{5}\right)\left(\frac{-3}{2}\right)$$ | Original expression. |
| $$\left(\frac{-4}{5}\right)\left(\frac{-3}{2}\right) = \frac{-4 \times -3}{5 \times 2}$$ | Multiply the numerators and the denominators separately. |
| $$\frac{-4 \times -3}{5 \times 2} = \frac{12}{10}$$ | Compute the products: $-4 \times -3 = 12$ and $5 \times 2 = 10$. |
| $$\frac{12}{10} = \frac{12 \div 2}{10 \div 2}$$ | Simplify the fraction by dividing both numerator and denominator by $2$. |
| $$\frac{12 \div 2}{10 \div 2} = \frac{6}{5}$$ | Final simplified result. |

</details>

## Question 2. Fraction Arithmetic.

### a. $$\dfrac{\left(\dfrac{3}{5}\right)}{\left(-\dfrac{7}{2}\right)}$$

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

| Equation | Explanation |
|--------------------------------------------------------------------------|------------------------------------------------------------------------------------------------------|
| $$\frac{\left(\frac{3}{5}\right)}{\left(-\frac{7}{2}\right)}$$ | Original expression. |
| $$ \frac{3}{5} \div \left(-\frac{7}{2}\right)$$ | Rewrite the complex fraction as a division problem. |
| $$\frac{3}{5} \times \left(-\frac{2}{7}\right)$$ | **Keep Change Flip.** Dividing by a fraction equals multiplying by its reciprocal; the reciprocal of $-\frac{7}{2}$ is $-\frac{2}{7}$. |
| $$\frac{3 \times (-2)}{5 \times 7}$$ | Multiply the numerators and denominators separately. |
| $$\frac{-6}{35}$$ | Compute the products: $3 \times (-2) = -6$ and $5 \times 7 = 35$. |
| $$\frac{-6}{35}$$ | Final simplified result. |

</details>

### b. $$\dfrac{-2}{\left(-\dfrac{4}{5}\right)}$$

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

| Equation | Explanation |
|---------------------------------------------------------------------|--------------------------------------------------------------------------------------------------|
| $$\frac{-2}{\left(-\frac{4}{5}\right)}$$ | Original expression. |
| $$ \dfrac{-2}{1} \div \left(-\frac{4}{5}\right)$$ | Rewrite the fraction as a division problem. |
| $$ \dfrac{-2}{1} \times \left(-\frac{5}{4}\right)$$ | **Keep Change Flip.** Dividing by a fraction equals multiplying by its reciprocal; the reciprocal of $-\frac{4}{5}$ is $-\frac{5}{4}$. |
| $$ \frac{-2 \times -5}{1 \times 4}$$ | Multiply the numerators and denominators. |
| $$ \frac{10}{4}$$ | Compute the product: $-2 \times -5 = 10$. |
| $$ \frac{5}{2}$$ | Simplify the fraction by dividing both numerator and denominator by $2$. |
| $$\frac{5}{2}$$ | Final simplified result. |

</details>

## Question 3. Fraction Arithmetic

### a. $$\dfrac{\left(-\dfrac{3}{4}\right)}{5}$$

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

| Equation | Explanation |
|--------------------------------------------------------------------------|--------------------------------------------------------------------------------------------------------------|
| $$\frac{\left(-\frac{3}{4}\right)}{5}$$ | Original expression. |
| $$ \left(-\frac{3}{4}\right) \div \dfrac{5}{1}$$ | Rewrite the complex fraction as a division problem. |
| $$ \left(-\frac{3}{4}\right) \times \frac{1}{5}$$ | Dividing by $5$ equals multiplying by its reciprocal; the reciprocal of $5$ is $\frac{1}{5}$. |
| $$ \frac{-3 \times 1}{4 \times 5}$$ | Multiply the numerators and denominators separately. |
| $$\frac{-3}{20}$$ | Compute the products: $-3 \times 1 = -3$ and $4 \times 5 = 20$. |
| $$\frac{-3}{20}$$ | Final simplified result. |

</details>

### b. $$\dfrac{2}{5}-\dfrac{3}{2}$$

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

| Equation | Explanation |
|-------------------------------------------------------|-------------------------------------------------------------------------------------------------------|
| $$\frac{2}{5}-\frac{3}{2}$$ | Original expression. |
| $$\frac{4}{10}-\frac{3}{2}$$ | Convert $\frac{2}{5}=\dfrac{4}{10}$ to an equivalent fraction with denominator $10$. |
| $$\frac{4}{10}-\frac{15}{10}$$ | Convert $\frac{3}{2}=\dfrac{15}{10}$ to an equivalent fraction with denominator $10$. | |
| $$ \frac{4-15}{10}$$ | Since the denominators are the same, subtract the numerators: $4-15$. |
| $$ \frac{-11}{10}$$ | Compute the subtraction in the numerator to obtain the final result, $\frac{-11}{10}$. |

</details>

## Question 4. Multiplication and Division by powers of 10.

### a. $$(3.276)(100,000)$$

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

| Equation | Explanation |
|------------------------------------------------------|-----------------------------------------------------------------------------------------------------|
| $$(3.276)(100,000)$$ | Original expression. |
| $$3.276 \times 100,000$$ | Recognize that multiplying by $100,000$ shifts the decimal point five places to the right. |
| $$3.276 \times 100,000 = 327600$$ | Perform the multiplication to obtain the final result, $327600$. |

</details>

### b. $$\dfrac{45.14}{10,000}$$

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

| Equation | Explanation |
|------------------------------------------------------------------|--------------------------------------------------------------------------------------------------------------|
| $$\frac{45.14}{10,000}$$ | Original expression. |
| $$45.14 \div 10,000 = 0.004514$$ | Dividing by $10,000$ shifts the decimal point four places to the left. |
| $$0.004514$$ | Final simplified result. |

</details>

## Question 5. Exponent Laws and Monomials.

### a. $$(4x^3)(2x^9)$$

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

| Equation | Explanation |
|------------------------------------------------------|-----------------------------------------------------------------------------------------------|
| $$(4x^3)(2x^9)$$ | Original expression. |
| $$ (4 \times 2)(x^3 \times x^9)$$ | Multiply the coefficients and the variable parts separately. |
| $$ 8x^{3+9}$$ | Compute $4 \times 2 = 8$ and apply the law of exponents: $x^3 \times x^9 = x^{3+9}$. |
| $$ 8x^{12}$$ | Simplify the exponent: $3+9 = 12$, yielding the final result. |

</details>

### b. $$\dfrac{16x^2}{4x^7}$$

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

| Equation | Explanation |
|-------------------------------------------------------|-------------------------------------------------------------------------------------------------------|
| $$\frac{16x^2}{4x^7}$$ | Original expression. |
| $$ \frac{16}{4} \cdot \frac{x^2}{x^7}$$| Separate the coefficients and the variable parts. |
| $$4x^{2-7}$$ | Divide the results to obtain the final expression. |
| $$4x^{-5}$$ | Divide the results to obtain the final expression. |

</details>

### c. $$4(x^2)^5$$

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

| Equation | Explanation |
|-------------------------------------------------------|-------------------------------------------------------------------------------------------------------|
| $$4(x^2)^5$$ | Original expression. |
| $$ 4 \cdot (x^2)^5$$ | Express the multiplication explicitly. |
| $$ 4 \cdot x^{2 \cdot 5}$$ | Apply the power-of-a-power rule: multiply the exponents. |
| $$4x^{10}$$ | Multiply the coefficient with the simplified variable part to get the final result. |

</details>

## Question 6. 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 7. 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 8. 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 9. 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 10. 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 11. 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 12. 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 13. 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 14. 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 15. 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 16. 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 17. 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 18. 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 19. 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 20. [Piecewise functions](https://hackmd.io/@m105collegealgebra/HJuz8ObJ1x) 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 21. 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 22. 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 23. 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 24. 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 25. 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 26. 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 27. 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 28. 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>