# Question 1:
Solve each inequality, use interval notation.

## a. $$6x - x^2 > 0$$

<details>
<summary> Example: </summary>

Solve $5x-x^2>0$

We are tasked with solving the inequality:

$$
5x - x^2 > 0
$$

### Step 1: Rewrite the inequality

First, rewrite the inequality as:

$$
x(5 - x) > 0
$$

Now, let's find the critical points by solving when the expression equals zero:

$$
x(5 - x) = 0
$$

The solutions are $x = 0$ and $x = 5$.

These points divide the number line into three intervals: $(-\infty, 0)$, $(0, 5)$, and $(5, \infty)$.

### Step 2: Create a sign chart

We will test points within each interval to determine the sign of the expression $x(5 - x)$.

| Interval | Test Point | Evaluate $x(5 - x)$ | Positive or Negative |
|----------------|------------|-------------------------|----------------------|
| $(-\infty, 0)$ | $x = -1$ | $(-1)(5 - (-1)) = -6$ | Negative |
| $(0, 5)$ | $x = 2$ | $(2)(5 - 2) = 6$ | Positive |
| $(5, \infty)$ | $x = 6$ | $(6)(5 - 6) = -6$ | Negative |

### Step 3: Determine the solution

From the sign chart, the expression is positive in the interval $(0, 5)$. The inequality is strict ($>$), so we exclude the boundary points $x = 0$ and $x = 5$.

### Final Answer

The solution to the inequality is:

$$
0 < x < 5
$$

The solution in interval notation is:

$$(0,5)$$

</details>

## b. $$3x^2 - \dfrac{x^3}{3} < 0$$

<details> <summary> Example: </summary>

We are tasked with solving:

$$
2x^2 - \frac{x^3}{4} < 0
$$

### Step 1: Multiply the inequality by 4

To eliminate the fraction, multiply both sides of the inequality by 4:

$$
4 \left(2x^2 - \frac{x^3}{4}\right) < 4 \cdot 0
$$

This simplifies to:

$$
8x^2 - x^3 < 0
$$

### Step 2: Factor the inequality

Factor the expression:

$$
x^2(8 - x) < 0
$$

### Step 3: Find the critical points

Set the expression equal to zero to find the critical points:

$$
x^2(8 - x) = 0
$$

The solutions are:

- $x^2 = 0 \quad \Rightarrow \quad x = 0$
- $8 - x = 0 \quad \Rightarrow \quad x = 8$

Thus, the critical points are $x = 0$ and $x = 8$. These points divide the number line into three intervals: $(-\infty, 0)$, $(0, 8)$, and $(8, \infty)$.

### Step 4: Create a sign chart

We will test points within each interval to determine the sign of the expression $x^2(8 - x)$.

| Interval | Test Point | Evaluate $x^2(8 - x)$ | Positive or Negative |
|----------------|------------|------------------------|----------------------|
| $(-\infty, 0)$ | $x = -1$ | $(-1)^2(8 - (-1)) = 9$ | Positive |
| $(0, 8)$ | $x = 4$ | $(4)^2(8 - 4) = 64$ | Positive |
| $(8, \infty)$ | $x = 10$ | $(10)^2(8 - 10) = -200$ | Negative |

### Step 5: Determine the solution

The expression $x^2(8 - x)$ is negative only in the interval $(8, \infty)$. Since the inequality is strict ($<$), we exclude the boundary points $x = 0$ and $x = 8$.

### Final Answer

The solution to the inequality is:

$$
x > 8
$$

The solution in interval notation is:

$$(-4,2)$$

</details>

## c. $$5 - x^2 > 2 - 2x$$

<details> <summary> Example: </summary>

$$2-x^2>10-7x$$

### Step 1: Move all terms to one side
First, we'll rearrange the inequality to bring all terms to one side:

$$2 - x^2 > 12 - 7x$$

This simplifies to:

$$-x^2 + 7x - 10 > 0$$

### Step 2: Factor the quadratic
We can factor out -1 to make factoring easier:

$$-(x^2 - 7x + 10) > 0$$

Now we need to factor the quadratic $x^2 - 7x + 10$:

$$x^2 - 7x + 10 = (x - 5)(x - 2)$$

Thus, the inequality becomes:

$$-(x - 5)(x - 2) > 0$$

### Step 3: Find the critical points
Set the factored expression equal to zero to find the critical points:

$$(x - 5)(x - 2) = 0$$

The critical points are:

$$x = 5$$
$$x = 2$$

These points divide the number line into three intervals: $(-∞, 2), (2, 5)$, and $(5, ∞)$.

### Step 4: Create a sign chart
We will test points within each interval to determine the sign of the expression $-(x - 5)(x - 2)$.

| Interval | Test Point | Evaluate -(x - 5)(x - 2) | Positive or Negative |
|----------------|------------|---------------------------|----------------------|
| $(-∞, 2)$ | $x = 0$ | $-((0 - 5)(0 - 2)) = -(-5)(-2) = -10$ | Negative |
| $(2, 5)$ | $x = 3$ | $-((3 - 5)(3 - 2)) = -(-2)(1) = 2$ | Positive |
| $(5, ∞)$ | $x = 6$ | $-((6 - 5)(6 - 2)) = -((1)(4)) = -4$ | Negative |

### Step 5: Determine the solution
We are looking for where the expression $-(x - 5)(x - 2)$ is greater than zero, which occurs in the interval $(2, 5)$.

### Final Answer
The solution to the inequality is:

$$2 < x < 5$$

The solution in interval notation is $(2, 5)$.

</details>

## d. $$x^2 + 1 < 2x + 9$$

<details> <summary> Example: </summary>

$$x^2+2<10-2x$$

We are tasked with solving the inequality:

$$
x^2 + 2 < 10 - 2x
$$

### Step 1: Move all terms to one side

First, move all terms to one side of the inequality:

$$
x^2 + 2 + 2x - 10 < 0
$$

Simplifying:

$$
x^2 + 2x - 8 < 0
$$

### Step 2: Factor the quadratic

We now factor the quadratic expression:

$$
x^2 + 2x - 8 = (x - 2)(x + 4)
$$

Thus, the inequality becomes:

$$
(x - 2)(x + 4) < 0
$$

### Step 3: Find the critical points

Set the factored expression equal to zero to find the critical points:

$$
(x - 2)(x + 4) = 0
$$

The critical points are:

- $x = 2$
- $x = -4$

These points divide the number line into three intervals: $(-\infty, -4)$, $(-4, 2)$, and $(2, \infty)$.

### Step 4: Create a sign chart

We will test points within each interval to determine the sign of the expression $(x - 2)(x + 4)$.

| Interval | Test Point | Evaluate $(x - 2)(x + 4)$ | Positive or Negative |
|----------------|------------|---------------------------|----------------------|
| $(-\infty, -4)$ | $x = -5$ | $(-5 - 2)(-5 + 4) = (-7)(-1) = 7$ | Positive |
| $(-4, 2)$ | $x = 0$ | $(0 - 2)(0 + 4) = (-2)(4) = -8$ | Negative |
| $(2, \infty)$ | $x = 3$ | $(3 - 2)(3 + 4) = (1)(7) = 7$ | Positive |

### Step 5: Determine the solution

We are looking for where the expression $(x - 2)(x + 4)$ is less than zero, which occurs in the interval $(-4, 2)$.

### Final Answer

The solution to the inequality is:

$$
-4 < x < 2
$$

The solution in interval notation is $$(-4,2)$$

</details>

---

# Question 2:
Find the following derivatives.

## a. For the function $$f(x) = e^{2x^2} - 5x$$ find: $$f'(x) =$$

<details> <summary> Example: </summary>

To find the derivative of the function:

$$
f(x) = e^{3x^3} - 4x
$$

we will apply the chain rule for the first term and the basic derivative rule for the second term.

### Step-by-Step Differentiation (Chain Rule for $e^{3x^3}$)

| | Outside Function | Inside Function |
|------------------------|-------------------------|----------------------|
| Original | $e^u$ | $u = 3x^3$ |
| Derivative | $e^u$ | $u' = 9x^2$ |

### Explanation

- The **outside function** is $e^u$, where $u = 3x^3$. The derivative of $e^u$ is $e^u$.
- The **inside function** is $3x^3$, and its derivative is $9x^2$.

Thus, applying the chain rule:

$$
\frac{d}{dx}(e^{3x^3}) = e^{3x^3} \cdot 9x^2
$$

### Full Derivative

For the second term $-4x$, its derivative is straightforward:

$$
\frac{d}{dx}(-4x) = -4
$$

### Final Answer

Thus, the derivative of the function is:

$$
f'(x) = 9x^2 e^{3x^3} - 4
$$

This result shows that for the term $e^{3x^3}$, we first take the derivative of the outside function $e^u$ and multiply it by the derivative of the inside function $3x^3$.

</details>

## b. For the function $$g(x) = -3x^4 + 2x^3 - 7$$ find: $$g'(x) =$$

<details> <summary> Example: </summary>

To find the derivative of the function:

$$
g(x) = -4x^3 + 5x^2 - 6
$$

we will apply the power rule to each term individually.

### Step-by-Step Differentiation

1. **First term: $-4x^3$**

The derivative of $-4x^3$ using the power rule ($\frac{d}{dx} x^n = nx^{n-1}$) is:

$$
\frac{d}{dx}(-4x^3) = -12x^2
$$

2. **Second term: $5x^2$**

The derivative of $5x^2$ is:

$$
\frac{d}{dx}(5x^2) = 10x
$$

3. **Third term: $-6$**

The derivative of the constant $-6$ is:

$$
\frac{d}{dx}(-6) = 0
$$

### Final Derivative

Thus, the derivative of the function is:

$$
g'(x) = -12x^2 + 10x
$$

</details>

## c. For the function $$h(x) = \frac{x}{x - 9}$$ find: $$h'(x) =$$

<details> <summary> Example: </summary>

To find the derivative of the function:

$$
h(x) = \frac{x}{x - 4}
$$

we will use the quotient rule, which states that if you have a function $h(x) = \frac{f(x)}{g(x)}$, the derivative is:

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

### Step-by-Step Differentiation

| | Top Function $f(x)$ | Bottom Function $g(x)$ |
|---|-----------------------|------------------------|
| Original | $x$ | $x - 4$ |
| Derivative | $f'(x) = 1$ | $g'(x) = 1$ |

### Applying the Quotient Rule

Using the quotient rule:

$$
h'(x) = \frac{(1)(x - 4) - (x)(1)}{(x - 4)^2}
$$

Simplify the numerator:

$$
h'(x) = \frac{x - 4 - x}{(x - 4)^2}
$$

$$
h'(x) = \frac{-4}{(x - 4)^2}
$$

### Final Answer

Thus, the derivative of the function is:

$$
h'(x) = \frac{-4}{(x - 4)^2}
$$

</details>

---

# Question 3:
Find the area bounded by the graphs of the indicated equations over the given interval.

## a. $$y = x^3 + 1; \hspace{2cm} y = 0; \hspace{2cm} 0 \leq x \leq 2$$

<details> <summary> Example: </summary>

To find the area bounded by the graphs of the equations:

$$
y = x^3 + 2 \quad \text{and} \quad y = 0 \quad \text{over the interval} \quad 0 \leq x \leq 3
$$

we need to compute the integral of $y = x^3 + 2$ from $x = 0$ to $x = 3$. Since $y = 0$ represents the x-axis, the area between the curves is simply the integral of $x^3 + 2$.


### Step-by-Step Solution

We will set up the definite integral:

$$
\text{Area} = \int_0^3 (x^3 + 2) \, dx
$$

### Step 1: Integrate the function

The integral of $x^3 + 2$ is:

$$
\int (x^3 + 2) \, dx = \frac{x^4}{4} + 2x
$$

### Step 2: Evaluate the integral from 0 to 3

Now, evaluate the integral at the limits:

\begin{align*}
\text{Area} &= \left[ \frac{x^4}{4} + 2x \right]_0^3 \\
&= \left( \frac{3^4}{4} + 2(3) \right) - \left( \frac{0^4}{4} + 2(0) \right) \\
&= \left( \frac{81}{4} + 6 \right) - 0 \\
&= \frac{81}{4} + \frac{24}{4} \\
&= \frac{105}{4}
\end{align*}

### Final Answer

The area bounded by the curves is:

$$
\text{Area} = \frac{105}{4} = 26.25 \, \text{square units}
$$


</details>

## b. $$y = -x^3 + 3;\hspace{2cm} y = 0; \hspace{2cm} -2 \leq x \leq 1$$

<details> <summary> Example: </summary>


To find the area bounded by the graphs of the equations:

$$
y = -x^3 + 4 \quad \text{and} \quad y = 0 \quad \text{over the interval} \quad -3 \leq x \leq 1
$$

we need to compute the integral of $y = -x^3 + 4$ from $x = -3$ to $x = 1$. Since $y = 0$ represents the x-axis, the area between the curves is simply the integral of $-x^3 + 4$.

### Step-by-Step Solution

We will set up the definite integral:

$$
\text{Area} = \int_{-3}^1 (-x^3 + 4) \, dx
$$

### Step 1: Integrate the function

The integral of $-x^3 + 4$ is:

$$
\int (-x^3 + 4) \, dx = -\frac{x^4}{4} + 4x
$$

### Step 2: Evaluate the integral from $-3$ to $1$

Now, evaluate the integral at the limits:

\begin{align*}
\text{Area} &= \left[ -\frac{x^4}{4} + 4x \right]_{-3}^1 \\
&= \left( -\frac{1^4}{4} + 4(1) \right) - \left( -\frac{(-3)^4}{4} + 4(-3) \right) \\
&= \left( -\frac{1}{4} + 4 \right) - \left( -\frac{81}{4} - 12 \right) \\
&= \left( \frac{16}{4} - \frac{1}{4} \right) - \left( -\frac{81}{4} - \frac{48}{4} \right) \\
&= \frac{15}{4} - \left( -\frac{129}{4} \right) \\
&= \frac{15}{4} + \frac{129}{4} \\
&= \frac{144}{4} \\
&= 36
\end{align*}

### Final Answer

The area bounded by the curves is:

$$
\text{Area} = 36 \, \text{square units}
$$


</details>

---

# Question 4: Use a definite integral to find the area bounded by the graph of the indicated equations over the given interval.
$$y = x^2 - 1; \hspace{1cm} y = x - 2; \hspace{1cm} -2 \leq x \leq 1$$

<details> <summary> Example: </summary>

Use a definite integral to find the area bounded by the graph of the indicated equations over the given interval.
$$y = x^2 -2; \hspace{1cm} y = x - 3; \hspace{1cm} -1 \leq x \leq 2$$


We are tasked with finding the area bounded by the curves:

$$
y = x^2 - 2 \quad \text{and} \quad y = x - 3 \quad \text{over the interval} \quad -1 \leq x \leq 2
$$

### Step 1: Determine the curves and find points of intersection

We need to find the points where the curves intersect. Set the two equations equal to each other:

$$
x^2 - 2 = x - 3
$$

Rearranging terms:

$$
x^2 - x + 1 = 0
$$

The discriminant of this quadratic equation is:

$$
\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3
$$

Since the discriminant is negative, there are no real solutions, which means the curves do not intersect within the interval. Therefore, we need to calculate the area between the curves by finding the integral of the difference between the two functions.

### Step 2: Set up the integral

To find the area between the curves, we integrate the difference between the top function and the bottom function over the interval $[-1, 2]$. Since $y = x - 3$ is above $y = x^2 - 2$ within the given interval, the area is:

$$
\text{Area} = \int_{-1}^{2} \left( (x^2 - 2)-(x - 3) \right) \, dx
$$

Simplify the integrand:

$$
\text{Area} = \int_{-1}^{2} \left( x^2-2-x + 3 \right) \, dx = \int_{-1}^{2} \left( x^2 - x + 1 \right) \, dx
$$

### Step 3: Integrate the function

Now we integrate term by term:

$$
\int \left( x^2 - x + 1 \right) \, dx = \frac{x^3}{3} - \frac{x^2}{2} + x
$$

### Step 4: Evaluate the integral from $x = -1$ to $x = 2$

Now, substitute the limits of integration:

\begin{align*}
\text{Area} &= \left[ \frac{x^3}{3} - \frac{x^2}{2} + x \right]_{-1}^{2} \\
&= \left( \frac{2^3}{3} - \frac{2^2}{2} + 2 \right) - \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} + (-1) \right) \\
&= \left( \frac{8}{3} - 2 + 2 \right) - \left( -\frac{1}{3} - \frac{1}{2} - 1 \right) \\
&= \left( \frac{8}{3} \right) - \left( -\frac{1}{3} -\frac{1}{2} - 1 \right)
\end{align*}

### Step 5: Simplify the evaluation

Simplifying each part:

$$
\left( \frac{8}{3} \right) + \left( \frac{1}{3} + \frac{1}{2} + 1 \right) = \frac{8}{3} + \left( \frac{1}{3} + \frac{3}{2} \right)
$$

Convert everything to a common denominator of 6:

$$
= \frac{8}{3} + \left( \frac{2}{6} + \frac{9}{6} \right) = \frac{8}{3} + \frac{11}{6}
$$

Find a common denominator:

$$
= \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = 4.5
$$

$$
\text{Area} = 4.5 \, \text{square units}
$$


</details>