# Question 1

Create a sign chart and determine where $f(x)>0$ in interval and inequality notation.

a. $f(x)=x^2-6x+8$

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<summary> Example: </summary>

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

### Step 1: Solve the equality $f(x) = 0$

We solve the quadratic equation $x^2 - 7x + 10 = 0$. Factoring gives:

$$
x^2 - 7x + 10 = (x - 5)(x - 2) = 0
$$

Thus, the solutions are:

$$
x = 5 \quad \text{and} \quad x = 2
$$

### Step 2: Sign chart and testing the sign of $f(x)$ in each interval

We use the critical points $x = 2$ and $x = 5$ to divide the real line into three intervals. Let’s test the sign of $f(x)$ within each interval.

| Interval | Test $x$ value | $f(x) = (x - 5)(x - 2)$ | Sign |
|------------------|----------------|--------------------------------|----------|
| $(-\infty, 2)$ | 0 | $f(0) = (0 - 5)(0 - 2) = 10$ | Positive |
| $(2, 5)$ | 3 | $f(3) = (3 - 5)(3 - 2) = -2$ | Negative |
| $(5, \infty)$ | 6 | $f(6) = (6 - 5)(6 - 2) = 4$ | Positive |

### Step 3: Solution to the inequality

We want to find where $f(x) > 0$, meaning the intervals where the function is positive. From the sign chart:

- $f(x) > 0$ in the intervals $(-\infty, 2)$ and $(5, \infty)$.

Thus, the solution in interval notation is:

$$
(-\infty, 2) \cup (5, \infty)
$$

The solution in inequality notation is:

$$x<2 \qquad \text{or} \qquad x>5$$

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b. $f(x)=x^2+4x-5$

::: spoiler
<summary> Example: </summary>

$f(x)=x^2+6x-7>0$

### Step 1: Solve the equality $f(x) = 0$

We solve the quadratic equation $x^2 + 6x - 7 = 0$. Factoring gives:

$$
x^2 + 6x - 7 = (x + 7)(x - 1) = 0
$$

Thus, the solutions are:

$$
x = -7 \quad \text{and} \quad x = 1
$$

### Step 2: Sign chart and testing the sign of $f(x)$ in each interval

We use the critical points $x = -7$ and $x = 1$ to divide the real line into three intervals. Let’s test the sign of $f(x)$ within each interval.

| Interval | Test $x$ value | $f(x) = (x + 7)(x - 1)$ | Sign |
|------------------|----------------|--------------------------------|----------|
| $(-\infty, -7)$ | -8 | $f(-8) = (-8 + 7)(-8 - 1) = 9$ | Positive |
| $(-7, 1)$ | 0 | $f(0) = (0 + 7)(0 - 1) = -7$ | Negative |
| $(1, \infty)$ | 2 | $f(2) = (2 + 7)(2 - 1) = 9$ | Positive |

### Step 3: Solution to the inequality

We want to find where $f(x) > 0$, meaning the intervals where the function is positive. From the sign chart:

- $f(x) > 0$ in the intervals $(-\infty, -7)$ and $(1, \infty)$.

Thus, the solution in interval notation is:

$$
(-\infty, -7) \cup (1, \infty)
$$

The solution in inequality notation is:

$$
x < -7 \qquad \text{or} \qquad x > 1
$$

:::

c. $f(x)=2x^2+9x-5>0$

::: spoiler
<summary> Example: </summary>

$3 x^2 - 19 x - 14>0$

### Step 1: Solve the equality $f(x) = 0$

We solve the quadratic equation $3x^2 - 19x - 14 = 0$. To factor this, the first term $3x^2$ splits as $(3x)(x)$. The last term $-14$ splits as $( \pm 14)( \mp 1), (\pm 1)(\mp 14), (\pm 7)(\mp 2), (\pm 2)(\mp 7)$

We check all the possibilities:

1. $(3x - 14)(x + 1) = 3x^2 - 11x - 14$

2. $(3x + 14)(x - 1) = 3x^2 + 11x - 14$

3. $(3x - 1)(x + 14) = 3x^2 + 41x - 14$

4. $(3x + 1)(x - 14) = 3x^2 - 41x - 14$

5. $(3x - 7)(x + 2) = 3x^2 - x - 14$

6. $(3x + 7)(x - 2) = 3x^2 + x - 14$

7. $(3x + 2)(x - 7) = 3x^2 - 19x - 14$

8. $(3x - 2)(x + 7) = 3x^2 + 19x - 14$

The 7th one is the correct factorization

$$
(3x + 2)(x - 7) = 0
$$

So, the solutions are:

$$
x = -\frac{2}{3} \quad \text{and} \quad x = 7
$$

### Step 2: Sign chart and testing the sign of $f(x)$ in each interval

We use the critical points $x = -\frac{2}{3}$ and $x = 7$ to divide the real line into three intervals. Let’s test the sign of $f(x)$ within each interval.

| Interval | Test $x$ value | $f(x) = (3x + 2)(x - 7)$ | Sign |
|----------------------|-----------------|-------------------------------------|----------|
| $(-\infty, -\frac{2}{3})$ | $-1$ | $f(-1) = (3(-1) + 2)((-1) - 7) = 32$| Positive |
| $(-\frac{2}{3}, 7)$ | $0$ | $f(0) = (3(0) + 2)(0 - 7) = -14$ | Negative |
| $(7, \infty)$ | $8$ | $f(8) = (3(8) + 2)(8 - 7) = 26$ | Positive |

### Step 3: Solution to the inequality

We want to find where $f(x) > 0$, meaning the intervals where the function is positive. From the sign chart:

- $f(x) > 0$ in the intervals $\left(-\infty, -\frac{2}{3}\right)$ and $(7, \infty)$.

Thus, the solution in interval notation is:

$$
\left(-\infty, -\frac{2}{3}\right) \cup (7, \infty)
$$

The solution in inequality notation is:

$$
x < -\frac{2}{3} \quad \text{or} \quad x > 7
$$

:::

# Question 3 Find the intervals on which $f(x)$ is increasing, decreasing, and local extrema.

a. $f(x)=5x^2-10x-3$

::: spoiler
<summary> Example: </summary>

$f(x)=7 x^2 + 28 x - 6$

### Step 1: Find the derivative of $f(x)$

The function is $f(x) = 7x^2 + 28x - 6$. Its derivative is:

$$
f'(x) = \frac{d}{dx}(7x^2 + 28x - 6) = 14x + 28
$$

### Step 2: Solve for critical points by setting $f'(x) = 0$

Set the derivative equal to zero:

$$
14x + 28 = 0
$$

Solve for $x$:

$$
x = -2
$$

So, the critical point is $x = -2$.

### Step 3: Sign chart for $f'(x)$

We will create a sign chart for $f'(x) = 14x + 28$ to determine where the function is increasing and decreasing.

| Interval | Test $x$ value | $f'(x) = 14x + 28$ | Sign |
|------------------|-----------------|--------------------------|----------|
| $(-\infty, -2)$ | $-3$ | $f'(-3) = 14(-3) + 28 = -14$ | Negative |
| $(-2, \infty)$ | $0$ | $f'(0) = 14(0) + 28 = 28$ | Positive |

### Step 4: Conclusion

- $f'(x) < 0$ on the interval $(-\infty, -2)$, so $f(x)$ is **decreasing** on $(-\infty, -2)$.
- $f'(x) > 0$ on the interval $(-2, \infty)$, so $f(x)$ is **increasing** on $(-2, \infty)$.
- There is a **local minimum** at $x = -2$ since the derivative changes from negative to positive.
- The $y$ value of the local minimum at $x=-2$ is found by plugging it into the original function. $f(-2)=7 (-2)^2 + 28 (-2) - 6=-34$.


:::

b. $f(x)=-x^3-2x-5$

::: spoiler
<summary> Example: </summary>

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

### Step 1: Find the derivative of $f(x)$

The function is $f(x) = -x^3 - 5x + 4$. Its derivative is:

$$
f'(x) = \frac{d}{dx}(-x^3 - 5x + 4) = -3x^2 - 5
$$

### Step 2: Solve for critical points by setting $f'(x) = 0$

Set the derivative equal to zero:

$$
-3x^2 - 5 = 0
$$

Solve for $x$:

$$
-3x^2 = 5 \quad \Rightarrow \quad x^2 = -\frac{5}{3}
$$

Since $x^2 = -\frac{5}{3}$ has no real solutions, there are no critical points.

### Step 3: Sign chart for $f'(x)$

Since the derivative has no real roots, the sign of $f'(x) = -3x^2 - 5$ is always negative. This means the function is always decreasing.

| Interval | Test $x$ value | $f'(x) = -3x^2 - 5$ | Sign |
|------------------|-----------------|-------------------------|----------|
| $(-\infty, \infty)$ | 0 | $f'(x) =-3(0)^2-5=-5$ | Negative |

### Step 4: Conclusion

- $f(x)$ is **decreasing** for all $x$: the interval $(-\infty, \infty)$.
- There are no local extrema because the function is always decreasing.

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