# Practice Exam 1

## 1. Find the following limits (if they exist) for $f(x)=\dfrac{2x^2-5x+2}{x^2+x-6}$.

:::spoiler
<summary>Simplification: </summary>
The top factors as:

$$2x^2-5x+2=(2x-1)(x-2)$$

The bottom factors as:

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

The fraction factors and simplifies to:

$$f(x)=\dfrac{2x^2-5x+2}{x^2+x-6}=\dfrac{(2x-1)(x-2)}{(x+3)(x-2)}=\dfrac{2x-1}{x+3}$$

Simplified:

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

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

:::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*}
:::
### (b) $\displaystyle \lim_{x \to 1} f(x)$

:::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{1}{4}
\end{align*}
:::

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

:::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{3}{5}
\end{align*}
:::

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

:::spoiler
<summary> Solution: </summary>

$\displaystyle \lim_{x \to -3} \frac{2x-1}{x+3}$

First, we recognize that direct substitution of $x = -3$ into the function $\frac{2x-1}{x+3}$ leads to an undefined expression due to division by zero. This indicates a potential discontinuity at $x = -3$. To better understand the behavior of the function as $x$ approaches -3, we will examine the limits from the left and right sides of -3.

#### Table of Values

To illustrate the behavior of the function near $x = -3$, let's consider values of $x$ approaching -3 from both the left (values less than -3) and the right (values greater than -3).

Here's the table with the calculated decimal y-values in the third column, while keeping the function expressions in the middle column:

| $x$ | $f(x)=\dfrac{2x-1}{x+3}$ | $\frac{2x-1}{x+3}$ (Decimal Value) |
|---------|------------------------------------|------------------------------------|
| -3.1 | $\frac{2(-3.1)-1}{-3.1+3}$ | 72.0 |
| -3.01 | $\frac{2(-3.01)-1}{-3.01+3}$ | 702.0 |
| -3.001 | $\frac{2(-3.001)-1}{-3.001+3}$ | 7002.0 |
| -2.9 | $\frac{2(-2.9)-1}{-2.9+3}$ | -68.0 |
| -2.99 | $\frac{2(-2.99)-1}{-2.99+3}$ | -698.0 |
| -2.999 | $\frac{2(-2.999)-1}{-2.999+3}$ | -6998.0 |

This table clearly illustrates how the function $\frac{2x-1}{x+3}$ diverges as $x$ approaches -3 from both directions, further emphasizing the non-existence of the limit at $x = -3$.

Calculating these values gives us an insight into the behavior of the function as $x$ approaches -3 from both sides.

#### Evaluating the Limits

As $x$ approaches -3 from the left, the values of $\frac{2x-1}{x+3}$ become increasingly positive, indicating that the function approaches positive infinity.

As $x$ approaches -3 from the right, the values of $\frac{2x-1}{x+3}$ become increasingly negative, indicating that the function approaches negative infinity.

#### Conclusion

The table of values indicates that as $x$ approaches -3, the function $\frac{2x-1}{x+3}$ does not approach a single finite value. Instead, it exhibits different behavior depending on the direction of approach:

- From the left ($x \to -3^-$), the function approaches positive infinity.
- From the right ($x \to -3^+$), the function approaches negative infinity.

Therefore, we conclude that $\displaystyle \lim_{x \to -3} \frac{2x-1}{x+3}$ does not exist because the left-hand limit and the right-hand limit are not equal and both approach infinity, albeit in opposite directions.

:::

## 2. Definition of a derivative

### (a) What is the definition of a derivative? In other words, given $f(x)$, the derivative of $f(x)$ at any point is given by the limit:

:::spoiler
<summary> Solution: </summary>
The derivative of

$$f(x)$$
is

$$f'(x)=\displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$$
:::

### (b) What information does the derivative give us?

:::spoiler
<summary> Solution: </summary>
The derivative tells us the slope of the tangent line of f(x) at any point x.
:::

### ( c) Find the derivative of $f(x)=x^2-2$ using the definition. That is, find the derivative using the 4-step process.

:::spoiler
<summary> Solution: </summary>

1.

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

2.

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

3.
\begin{align*}
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{2xh+h^2}{h} \\
&=\dfrac{2xh}{h}+\dfrac{h^2}{h} \\
&=2x+h
\end{align*}

4.
\begin{align*}
\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}&=\lim_{h \to 0}(2x+h) \\
&=2x+0 \\
&=2x
\end{align*}

Thus, the derivative is $$f'(x)=2x$$
:::

## 3. Refer to the function $g(x)$ to answer the following questions.

$g(x)=\begin{cases} \dfrac{x^2-4}{x-2} & x<2 \\ 0 & x=2 \\ x^2 & x>2 \end{cases}$

:::spoiler
<summary> Simplified g(x): </summary>

Simplify the piece where x is less than 2 by factoring and canceling (x-2).

$$\dfrac{x^2-4}{x-2}=\dfrac{(x+2)(x-2)}{x-2}=x+2$$

Simplified $g(x)=\begin{cases} x+2 & x<2 \\ 0 & x=2 \\ x^2 & x>2 \end{cases}$

:::

### (a) Find $\displaystyle \lim_{x \to 2^-} g(x)$

:::spoiler
<summary> Solution: </summary>

If $x \to 2^-$, that means x is approaching 2 from the left, or when $x<2$. Therefore, we use $x+2$ to plug 2 into.

$$\displaystyle \lim_{x \to 2^-} g(x)=\lim_{x \to 2^-} (x+2)=2+2=4$$

:::

### (b) Find $\displaystyle \lim_{x \to 2^+} g(x)$

:::spoiler
<summary> Solution: </summary>

If $x \to 2^+$, that means x is approaching 2 from the right, or when $x>2$. Therefore, we use $x^2$ to plug 2 into.

$$\displaystyle \lim_{x \to 2^+} g(x)=\lim_{x \to 2^+} x^2=2^2=4$$

:::

### ( c) Find $g(2)$

:::spoiler
<summary> Solution: </summary>

$g(2)$ means $x=2$ so we use the middle piece, 0.

$g(2)=0$.

:::

### (d) Is $g(x)$ continuous at $x=2$? Fully justify your answer.

:::spoiler
<summary> Solution: </summary>

1. $\displaystyle \lim_{x \to 2} g(x)=4$ since the limits from the left and right both equaled 4.

2. $g(2)=0$

3. Since $\displaystyle \lim_{x \to 2}g(x)=4\neq g(2)=0$ which fails the last condition for continuity at 2, making g(x) discontinuous at x=2.

:::

## 4. Find the equation of the tangent line to the function $f(x)=x^2-4x+7$ at $x=1$.

:::spoiler
<summary> Solution: </summary>

To find the equation of any line we need the slope of the line $m$, and any point that it passes through $(x_1,y_1)$, and use point-slope form of a line equation: $$y-y_1=m(x-x_1)$$

Since the function and its tangent line at $x=1$ also intersect at $x=1$, the $y$ value is the same as the $y$-value of the function.

Thus the $x$ value is $x_1=1$ and the y-value is found by plugging the x-value back into the original function: $y_1=f(1)=1^2-4(1)+7=1-4+7=4$, or $y_1=4$.

Now we have our point $(x_1,y_1)=(1,4)$.

To find the slope of the tangent line, we need the derivative $f'(x)$.

Taking the derivative using the power rule $\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}$:

\begin{align*}
f'(x)&=\dfrac{d}{dx} \left(x^2\right) -\dfrac{d}{dx} \left(4x\right)+\dfrac{d}{dx} \left(7\right) \\
&=\dfrac{d}{dx} \left(x^2\right) -\dfrac{d}{dx} \left(4x^1\right)+\dfrac{d}{dx} \left(7x^0\right) \\
&=2x^{2-1}-4(1x^{1-1})+7(0x^{0-1}) \\
f'(x)&=2x-4
\end{align*}

Since we want to find the slope of the tangent line of $f(x)$ at $x=1$, we plug $x=1$ into the derivative.

$f'(1)=2(1)-4=2-4=-2$.

Thus our slope of our tangent line is $m=-2$ and it passes through the point $(x_1,y_1)=(1,4)$.

So our equation of the tangent line at $x=1$ is:

$$y-4=-2(x-1)$$
:::

## 5. What does it mean for a function to be continuous?

:::spoiler
<summary> Solution: </summary>

$f(x)$ is continuous at $x=a$ if all the three conditions are true:

1. $f(a)$ exists.
2. $\displaystyle \lim_{x \to a} f(x)$ exists
3. $\displaystyle \lim_{x \to a} f(x)=f(a)$

If $f(x)$ is continuous on an interval $[a,b]$ then it is continuous for every $x$ value in the interval.
:::

## 6. The manager of a furniture factory finds that it costs $2200 to manufacture 100 chairs in one day and $4800 to manufacture 300 chairs in one day. If C is a linear function that gives the cost to manufacture x number of chairs in one day, what is the slope of C and what does it represent in this context?

:::spoiler
<summary> Solution: </summary>

$$m=\dfrac{C_2-C_1}{x_2-x_1}=\dfrac{4800-2200}{300-100}=\dfrac{2600}{200}=\dfrac{13 \text{ dollars}}{1 \text{ chair}}$$

The cost of each additional chair is \$13 a chair.
:::

## 7. Rewrite the expression $\dfrac{\sqrt{x} \cdot x^2}{x^{1/3}}$

:::spoiler
<summary> Solution: </summary>

$x^n \cdot x^m=x^{n+m}$

$\dfrac{x^n}{x^m}=x^{n-m}$

$\sqrt{x}=x^{1/2}$

\begin{align*}
\dfrac{\sqrt{x} \cdot x^2}{x^{1/3}}&=\dfrac{x^{1/2} \cdot x^{2}}{x^{1/3}} \\
&=x^{1/2+2-1/3} \\
&=x^{\tfrac{3}{6}+\tfrac{12}{6}-\tfrac{2}{6}} \\
&=x^{\tfrac{3+12-2}{6}} \\
&=x^{\tfrac{13}{6}}
\end{align*}
:::

## 8. The following four questions all apply to the function $f(x)=x^2-4x$.

### (a) Find the average rate of change of $f(x)$ on the interval $[1,2]$.

:::spoiler
<summary> Solution: </summary>

| $x$ | $y=f(x)=x^2-4x$ |
| -------- | -------- |
| $1$ | $1^2-4(1)=-3$ |
| $2$ | $2^2-4(2)=-4$ |

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-4-(-3)}{2-1}=\dfrac{-1}{1}=-1$

:::

### (b) Find $f'(x)$.

:::spoiler
<summary> Solution: </summary>

\begin{align*}
f'(x)&=(x^2)'-(4x)' \\
&=2x^{2-1}-4(1x^{1-1}) \\
&=2x-4
\end{align*}

:::

### ( c) Find the instantaneous rate of change of $f(x)$ at $x=1$.

:::spoiler
<summary> Solution: </summary>

Plug $x=1$ into the derivative.

\begin{align*}
f'(x)&=2x-4 \\
f'(1)&=2(1)-4=-2
\end{align*}

:::

### (d) Find the tangent line to the graph of $f(x)$ at $x=1$.

:::spoiler
<summary> Solution: </summary>

$m=-2$ and $(x_1,y_1)=(1,-3)$ from parts a and b, so:

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

$$y-(-3)=-2(x-1)$$

$$y+3=-2(x-1)$$
:::

## 9 . The price $P$ a company can charge for DVD players is related to the number of players $q$ that are in demand by the public by the equation $P (q) = 75e^{−0.05q}$, where $q$ is in thousands and $P$ is in dollars. What does the statement $P'(10) = −2.27$ mean in this context?

:::spoiler
<summary> Solution: </summary>
When 10,000 DVD players are sold, the price is decreasing at a rate of $ 2.27 per additional 1000 DVD players sold
:::

## 10. Find the derivative for each of the following functions. Do not use the 4-step process. Use the rules for derivatives.

### (a) $f(x)=3x^4+\dfrac{2}{x^2}-\sqrt{x}$

:::spoiler
<summary> Solution: </summary>

Rewrite $f(x)=3x^4+\dfrac{2}{x^2}-\sqrt{x}$

$=3x^4+2x^{-2}-x^{1/2}$

Derivative: $f'(x)=3(4x^{4-1})+2(-2x^{-2-1})-(1/2)x^{1/2-1}$

$f'(x)=12x^3-4x^{-3}-\dfrac{1}{2}x^{-1/2}$

Simplified Derivative: $f'(x)=12x^3-\dfrac{4}{x^3}-\dfrac{1}{2x^{1/2}}$

:::

### (b) $K(x)=(3x+5)(x+2)$

:::spoiler
<summary> Solution: </summary>

Rewrite $K(x)=(3x+5)(x+2)$

$=3x^2+6x+5x+10$

$=3x^2+11x+10$

Derivative: $K'(x)=3(2x^{2-1})+11(1x^{1-1})+10(0x^{0-1})$

$K'(x)=6x+11$

:::
### ( c) $f(x)=\dfrac{x^2-5x+3}{\sqrt{x}}$

:::spoiler
<summary> Solution: </summary>

Rewrite

\begin{align*}
f(x)&=\dfrac{x^2-5x+3}{\sqrt{x}} \\
&=\dfrac{x^2-5x+3}{x^{1/2}} \\
&=\dfrac{x^2}{x^{1/2}}-\dfrac{5x}{x^{1/2}}+\dfrac{3}{x^{1/2}} \\
&=x^{2-1/2}-5x^{1-1/2}+3x^{-1/2} \\
f(x)&=x^{3/2}-5x^{1/2}+3x^{-1/2} \\
\end{align*}

Derivative:

\begin{align*}
f'(x)&=(x^{3/2}-5x^{1/2}+3x^{-1/2})' \\
&=(x^{3/2})'-(5x^{1/2})'+(3x^{-1/2})' \\
&=\dfrac{3}{2}x^{3/2-1} - 5 (1/2)x^{1/2-1} + 3(-1/2)x^{-1/2-1} \\
&=\dfrac{3}{2}x^{1/2}-\dfrac{5}{2}x^{-1/2} -\dfrac{3}{2}x^{-3/2} \\
f'(x)&=\dfrac{3}{2}x^{1/2}-\dfrac{5}{2x^{1/2}} -\dfrac{3}{2x^{3/2}} \\
\end{align*}

:::

## 11. The graph of a function $f(x)$ is shown below. Use the graph to answer the following questions.


:::spoiler

<summary> Where to look to find each Limit:</summary>

:::

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

:::spoiler
<summary> Solution: </summary>

As $x$ approaches -2 from the right, $f(x)$ approaches $-\infty$ (down forever).

The answer is $-\infty$.
:::

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

:::spoiler
<summary> Solution: </summary>

As $x$ approaches -2 from the left, $f(x)$ approaches $\infty$ (up forever).

The answer is $\infty$.

:::

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

:::spoiler
<summary> Solution: </summary>

As $x$ approaches 0 from both the left and right sides, $f(x)$ approaches $2$. (Look at the point $(0,2)$)

The answer is 2.

:::

### (d) $\displaystyle \lim_{x \to -1^+} f(x)$

:::spoiler
<summary> Solution: </summary>

As $x$ approaches -1 from the right side, $f(x)$ approaches $1$. (Look at the point $(-1,1)$)

The answer is 1.

:::

### (e) $\displaystyle \lim_{x \to -1^-} f(x)$

:::spoiler
<summary> Solution: </summary>

As $x$ approaches -1 from the left side, $f(x)$ approaches $0$. (Look at the point $(-1,0)$)

The answer is 0.

:::

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

:::spoiler
<summary> Solution: </summary>

Since the limit as x approaches -1 from the left (0) is different from the limit as x approaches -1 from the right (1), the limit as x approaches -1 does not exist.

The answer is the limit does not exist.

:::

### (g) At what $x$-values is $f$ not continuous?

:::spoiler
<summary> Solution: </summary>

The function is discontinuous at $x=-2$ since $g(-2)$ doesn't exist, failing the first condition for continuous at -2. (Looking at the graph it has a vertical asymptote at $x=-2$).

Also, the function is discontinuous at $x=-1$ since the limits from the left and right are not equal (see the previous part, f).

:::

# Comprehensive Companion Guide for Practice Exam 1

## Introduction

This guide aims to assist students in preparing for Practice Exam 1, focusing on essential calculus concepts such as limits, derivatives, continuity, and their applications, without delving into the product rule, quotient rule, intermediate value theorem, or chain rule. The document provides strategies, insights, and tips to improve understanding and performance.

## Section 1: Mastering Limits

### Conceptual Overview
Limits are foundational to calculus, defining the behavior of functions as inputs approach a specific value. They are key to understanding both the continuity of functions and the concept of derivatives.

### Detailed Study Tips
1. **Direct Substitution**: Start with direct substitution to solve limits when possible. This straightforward method works well unless it results in an indeterminate form like 0/0.
2. **Factoring and Simplifying**: For expressions that don't resolve easily, look for opportunities to factor polynomials and cancel terms to simplify the limit expression.
3. **Graphical Understanding**: Visualizing functions using graphs can provide intuitive insights into the behavior of limits, helping to predict whether a limit exists and its value.

### Common Pitfalls and How to Avoid Them
- Ignoring one-sided limits. Always check limits from both the left and right sides to ensure they agree for the limit to exist at a point.

## Section 2: Understanding Derivatives

### Conceptual Overview
Derivatives represent the rate of change of a function with respect to its input. They are crucial for analyzing the function's behavior, such as identifying rates of change and slopes of tangent lines.

### Detailed Study Tips
1. **Power Rule Mastery**: Focus on the power rule for differentiation, as it is a fundamental tool for finding derivatives, especially when the product rule, quotient rule, and chain rule are excluded.
2. **Higher-Order Derivatives**: Understand how to find second derivatives to explore acceleration and the concavity of functions.
3. **Application to Tangent Lines**: Practice finding the slope of tangent lines using derivatives, as this concept is pivotal for interpreting the behavior of functions graphically.

### Common Pitfalls and How to Avoid Them
- Forgetting to apply the power rule correctly, especially with negative and fractional exponents. Practice with a variety of functions to build confidence.

## Section 3: Continuity and Its Implications

### Conceptual Overview
Continuity at a point means a function is smooth and unbroken at that point, which is essential for applying the basic theorems of calculus.

### Detailed Study Tips
1. **Checking Continuity**: Learn to check for continuity by ensuring a function is defined at the point, the limit exists at that point, and the limit equals the function’s value.
2. **Discontinuities Identification**: Be able to identify different types of discontinuities and understand their implications on a function's behavior.

### Common Pitfalls and How to Avoid Them
- Assuming a function is continuous across its domain without verification. Always check for continuity at points of interest, especially where the function’s formula changes.

## Section 4: Calculus in Action - Application Problems

### Conceptual Overview
Calculus is a powerful tool for modeling real-world problems, enabling the analysis and optimization of systems in engineering, physics, economics, and beyond.

### Detailed Study Tips
1. **Understanding Problem Context**: Take time to understand the context of application problems, translating real-world scenarios into mathematical models.
2. **Solving for Real-World Variables**: Be adept at using derivatives to find maximums, minimums, and optimal solutions in applied contexts.

### Common Pitfalls and How to Avoid Them
- Misinterpreting the problem statement. Ensure you understand what is being asked, especially the relationships between different quantities in application problems.

## Exam Preparation Strategies

1. **Diverse Problem Solving**: Engage with a wide range of problems to strengthen your understanding and adaptability to different question types.
2. **Structured Review Sessions**: Organize your study time to cover all topics, dedicating specific sessions to each area for focused review.
3. **Peer Discussions**: Explaining concepts to others and solving problems together can deepen understanding and reveal new perspectives.
4. **Resource Utilization**: Make use of textbooks, online resources, and past exam papers for additional practice and clarification of concepts.
5. **Instructor Consultation**: Don't hesitate to seek clarification from your instructor on confusing topics or to gain deeper insights into complex concepts.
6. **Healthy Study Habits**: Ensure to balance study with breaks, maintain a healthy diet, and get adequate sleep to keep your mind sharp and ready for the exam.

## Conclusion

By focusing on these core areas and employing effective study and exam strategies, students can build a solid foundation in calculus principles and techniques. This guide aims to support your preparation for Practice Exam 1, fostering a deeper understanding and enhancing your ability to successfully tackle the exam and beyond.