---
# System prepended metadata

title: 4.5 Pearson Online HW Solutions
tags: [pearson, '4.5']

---

# 4.5 Pearson Online HW Solutions

## 1. By inspecting the graph of the function, find the absolute maximum and absolute minimum on the given interval. $q(x)=x^2+2$ on $[-3,2]$

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



    
![image](https://hackmd.io/_uploads/BJktFmiJkg.png)

    
### (On interval $[-3,2]$)
The function $f(x)$ has an absolute maximum value at the left endpoint $x=-3$ of $y$ value $y=11$.
    
The function $f(x)$ has an absolute minimum value at the critical value in between, $x=0$ of $y$ value $y=2$.

</details>

## 2. For the graph of $y=f(x)$ shown to the right, find the absolute minimum and the absolute maximum over the interval $[1,10]$.

![{4DB32A94-38BA-4768-9E14-1A702089F65D}](https://hackmd.io/_uploads/Bk8pdSvJke.png)

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

- On the interval $[1,10]$, the function $f(x)$ has an absolute maximum at the critical number in between, at $x=5$ of $y$ value $y=9$.
    
- On the interval $[1,10]$, the function $f(x)$ has an absolute minimum at the left endpoint, $x=1$ of $y$ value $y=2$.

</details>

## 3. For the graph of $y=f(x)$ shown to the right, find the absolute minimum and the absolute maximum over the interval $[1,5]$.


![{B6F43835-05C8-4756-93BE-4EAA79C5095F}](https://hackmd.io/_uploads/rkw0OrPkkg.png)


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

- On the interval $[1,5]$, the function $f(x)$ has an absolute maximum at the critical number in between, at $x=3$ of $y$ value $y=7$.
    
- On the interval $[1,5]$, the function $f(x)$ has an absolute minimum at the left endpoint, $x=0$ of $y$ value $y=2$.

</details>

## 4. Find the absolute maximum and minimum values of the function over the indicated interval. $$f(x)=2x^2-1$$

### a. $[2,7]$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = 2x^2 - 1
$$

We need to find the absolute maximum and absolute minimum of this function on the interval $[2, 7]$.

### Step 2: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(2x^2 - 1)
$$

The derivative is:

$$
f'(x) = 4x
$$

Now, set the derivative equal to zero to find the critical points:

$$
4x = 0
$$

Solve for $x$:

$$
x = 0
$$

Since $x = 0$ is not within the interval $[2, 7]$, there are no critical points within this interval.

### Step 3: Evaluate the Function at the Endpoints

Since there are no critical points in the interval, we only need to evaluate the function $f(x)$ at the endpoints of the interval $x = 2$ and $x = 7$.

- At $x = 2$:

$$
f(2) = 2(2)^2 - 1 = 2(4) - 1 = 8 - 1 = 7
$$

- At $x = 7$:

$$
f(7) = 2(7)^2 - 1 = 2(49) - 1 = 98 - 1 = 97
$$

### Step 4: Determine the Absolute Maximum and Minimum

Now, compare the values at the endpoints:

- $f(2) = 7$
- $f(7) = 97$

- The **absolute minimum** is $f(2) = 7$.
- The **absolute maximum** is $f(7) = 97$.

### Step 5: Verify with a Graph

We will now plot the function to visually verify the absolute maximum and minimum on the interval $[2, 7]$.

    
![image](https://hackmd.io/_uploads/SJEQiXiyye.png)


</details>

### b. $[-7,7]$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = 2x^2 - 1
$$

We need to find the absolute maximum and absolute minimum of this function on the interval $[-7, 7]$.

### Step 2: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(2x^2 - 1)
$$

The derivative is:

$$
f'(x) = 4x
$$

Now, set the derivative equal to zero to find the critical points:

$$
4x = 0
$$

Solve for $x$:

$$
x = 0
$$

Thus, $x = 0$ is a critical point within the interval $[-7, 7]$.

### Step 3: Evaluate the Function at the Critical Point and Endpoints

We will now evaluate the function $f(x)$ at the critical point $x = 0$ and at the endpoints of the interval $x = -7$ and $x = 7$.

- At $x = 0$:

$$
f(0) = 2(0)^2 - 1 = -1
$$

- At $x = -7$:

$$
f(-7) = 2(-7)^2 - 1 = 2(49) - 1 = 98 - 1 = 97
$$

- At $x = 7$:

$$
f(7) = 2(7)^2 - 1 = 2(49) - 1 = 98 - 1 = 97
$$

### Step 4: Determine the Absolute Maximum and Minimum

Now, compare the values at the critical point and the endpoints:

- $f(0) = -1$
- $f(-7) = 97$
- $f(7) = 97$

- The **absolute minimum** is $f(0) = -1$.
- The **absolute maximum** is $f(-7) = f(7) = 97$.

### Step 5: Verify with a Graph

We will now plot the function to visually verify the absolute maximum and minimum on the interval $[-7, 7]$.

![image](https://hackmd.io/_uploads/Hyk5jmskJx.png)

    
</details>

## 5. Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x-values at which they occur. $$f(x)=x^2-6x-9; \qquad [-1,6]$$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x^2 - 6x - 9
$$

We need to find the absolute maximum and absolute minimum of this function on the interval $[-1, 6]$.

### Step 2: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(x^2 - 6x - 9)
$$

The derivative is:

$$
f'(x) = 2x - 6
$$

Now, set the derivative equal to zero to find the critical points:

$$
2x - 6 = 0
$$

Solve for $x$:

$$
x = 3
$$

Thus, $x = 3$ is a critical point within the interval $[-1, 6]$.

### Step 3: Evaluate the Function at the Critical Point and Endpoints

We will evaluate the function $f(x)$ at the critical point $x = 3$ and at the endpoints of the interval $x = -1$ and $x = 6$.

### Step 4: Determine the Absolute Maximum and Minimum

We can summarize the calculations in the table below:

| $x$   | Work                                           | $f(x)$  |
|-------|------------------------------------------------|---------|
| $-1$  | $f(-1) = (-1)^2 - 6(-1) - 9 = 1 + 6 - 9$       | $-2$    |
| $3$   | $f(3) = (3)^2 - 6(3) - 9 = 9 - 18 - 9$         | $-18$   |
| $6$   | $f(6) = (6)^2 - 6(6) - 9 = 36 - 36 - 9$        | $-9$    |

- The **absolute minimum** value is $f(3) = -18$ at $x = 3$.
- The **absolute maximum** value is $f(-1) = -2$ at $x = -1$.

### Final Answer:

- The absolute minimum value is $-18$ at $x = 3$.
- The absolute maximum value is $-2$ at $x = -1$.


</details>

## 6. Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x-values at which they occur. $$f(x)=x^2-10x-1 ; \qquad [1,6]$$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x^2 - 10x - 1
$$

We need to find the absolute maximum and absolute minimum of this function on the interval $[1, 6]$.

### Step 2: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(x^2 - 10x - 1)
$$

The derivative is:

$$
f'(x) = 2x - 10
$$

Now, set the derivative equal to zero to find the critical points:

$$
2x - 10 = 0
$$

Solve for $x$:

$$
x = 5
$$

Thus, $x = 5$ is a critical point within the interval $[1, 6]$.

### Step 3: Evaluate the Function at the Critical Point and Endpoints

We will now evaluate the function $f(x)$ at the critical point $x = 5$ and at the endpoints of the interval $x = 1$ and $x = 6$.

### Step 4: Determine the Absolute Maximum and Minimum

We can summarize the calculations in the table below:

| $x$   | Work                                           | $f(x)$  |
|-------|------------------------------------------------|---------|
| $1$   | $f(1) = (1)^2 - 10(1) - 1 = 1 - 10 - 1$        | $-10$   |
| $5$   | $f(5) = (5)^2 - 10(5) - 1 = 25 - 50 - 1$       | $-26$   |
| $6$   | $f(6) = (6)^2 - 10(6) - 1 = 36 - 60 - 1$       | $-25$   |

- The **absolute minimum** value is $f(5) = -26$ at $x = 5$.
- The **absolute maximum** value is $f(1) = -10$ at $x = 1$.

### Final Answer:

- The absolute minimum value is $-26$ at $x = 5$.
- The absolute maximum value is $-10$ at $x = 1$.


</details>

## 7. Find the absolute extremum, if any, for the following function. $$f(x)=6x^4-1$$

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

### Step 1: Function

We are given the function:

$$
f(x) = 6x^4 - 1
$$

We need to find the absolute maximum and absolute minimum (if any) of this function.

### Step 2: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(6x^4 - 1)
$$

The derivative is:

$$
f'(x) = 24x^3
$$

Now, set the derivative equal to zero to find the critical points:

$$
24x^3 = 0
$$

Solve for $x$:

$$
x = 0
$$

Thus, $x = 0$ is a critical point.

### Step 3: Evaluate the Function at the Critical Point and Endpoints

Since the function is defined for all $x$, we do not have any specific endpoints. We will evaluate the function at the critical point $x = 0$ and analyze the behavior of the function as $x \to \infty$ and $x \to -\infty$.

- At $x = 0$:

$$
f(0) = 6(0)^4 - 1 = -1
$$

### Step 4: Analyze Behavior at Infinity

As $x \to \infty$, the term $6x^4$ dominates, and since $x^4$ is always positive for large positive or negative $x$, the function grows without bound:

$$
\lim_{x \to \infty} f(x) = \infty
$$

$$
\lim_{x \to -\infty} f(x) = \infty
$$

### Step 5: Determine the Absolute Extremum

We can summarize the analysis as follows:

| $x$             | Work                                  | $f(x)$  |
|-----------------|---------------------------------------|---------|
| $0$             | $f(0) = 6(0)^4 - 1 = -1$              | $-1$    |
| $x \to \infty$  | Function tends to infinity            | $\infty$|
| $x \to -\infty$ | Function tends to infinity            | $\infty$|

- The **absolute minimum** is $f(0) = -1$ at $x = 0$.
- The function increases without bound as $x \to \infty$ and $x \to -\infty$, so there is no absolute maximum.

### Final Answer:

- The absolute minimum value is $-1$ at $x = 0$.
- There is no absolute maximum.


</details>

## 8. Find the absolute minimum value on $[0,\infty)$ for $f(x)=2x^2-8x+9$.

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


### Step 1: Function and Interval

We are given the function:

$$
f(x) = 2x^2 - 8x + 9
$$

We need to find the absolute minimum value of this function on the interval $[0, \infty)$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(2x^2 - 8x + 9)
$$

This gives:

$$
f'(x) = 4x - 8
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
4x - 8 = 0
$$

Solve for $x$:

$$
4x = 8
$$

$$
x = 2
$$

### Step 4: Evaluate the Function at the Critical Point and Endpoint

We will evaluate the function $f(x)$ at the critical point $x = 2$ and at the endpoint $x = 0$.

- At $x = 2$:

$$
f(2) = 2(2)^2 - 8(2) + 9 = 8 - 16 + 9 = 1
$$

- At $x = 0$:

$$
f(0) = 2(0)^2 - 8(0) + 9 = 9
$$

### Step 5: Analyze Behavior at Infinity

As $x \to \infty$, the term $2x^2$ dominates the behavior of the function. Therefore, as $x \to \infty$:

$$
f(x) \to \infty
$$

### Step 6: Determine the Absolute Minimum

We can summarize the calculations as follows:

| $x$          | Work                                | $f(x)$  |
|--------------|-------------------------------------|---------|
| $0$          | $f(0) = 2(0)^2 - 8(0) + 9$          | $9$     |
| $2$          | $f(2) = 2(2)^2 - 8(2) + 9$          | $1$     |
| $x \to \infty$ | Function tends to infinity          | $\infty$|


- The **absolute minimum** value is $1$ at $x = 2$.
- The function increases without bound as $x \to \infty$, so there is no minimum at infinity.

### Final Answer:

- The absolute minimum value is $1$ at $x = 2$.


</details>

## 9. Find the absolute minimum value on $[0,\infty)$ for $f(x)=(x+14)(x-7)^2$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = (x+14)(x-7)^2
$$

We need to find the absolute minimum value of this function on the interval $[0, \infty)$.

### Step 2: Expand the Function

First, let's expand the function to make it easier to differentiate:

$$
f(x) = (x + 14)(x^2 - 14x + 49)
$$

Expand:

$$
f(x) = x(x^2 - 14x + 49) + 14(x^2 - 14x + 49)
$$

Simplify:

$$
f(x) = x^3 - 14x^2 + 49x + 14x^2 - 196x + 686
$$

Combine like terms:

$$
f(x) = x^3 + 0x^2 - 147x + 686
$$

Thus, the function is:

$$
f(x) = x^3 - 147x + 686
$$

### Step 3: Find the Critical Points

To find the critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}(x^3 - 147x + 686)
$$

The derivative is:

$$
f'(x) = 3x^2 - 147
$$

Now, set the derivative equal to zero to find the critical points:

$$
3x^2 - 147 = 0
$$

Solve for $x$:

$$
x^2 = \frac{147}{3} = 49
$$

$$
x = \pm 7
$$

However, since we are only interested in the interval $[0, \infty)$, we discard $x = -7$ and keep $x = 7$.

### Step 4: Evaluate the Function at the Critical Point and Endpoint

We will evaluate the function $f(x)$ at the critical point $x = 7$ and at the endpoint $x = 0$.

- At $x = 7$:

$$
f(7) = (7 + 14)(7 - 7)^2 = 21(0) = 0
$$

- At $x = 0$:

$$
f(0) = (0 + 14)(0 - 7)^2 = 14(49) = 686
$$

### Step 5: Analyze Behavior at Infinity

As $x \to \infty$, the term $x^3$ dominates the behavior of the function. Therefore, as $x \to \infty$:

$$
f(x) \to \infty
$$

This means that the function grows without bound as $x \to \infty$.

### Step 6: Determine the Absolute Minimum

We can summarize the calculations as follows:

| $x$   | Work                                           | $f(x)$  |
|-------|------------------------------------------------|---------|
| $0$   | $f(0) = (0 + 14)(0 - 7)^2 = 14(49)$            | $686$   |
| $7$   | $f(7) = (7 + 14)(7 - 7)^2 = 21(0)$             | $0$     |
| $x \to \infty$  | Function tends to infinity            | $\infty$|

- The **absolute minimum** value is $f(7) = 0$ at $x = 7$.
- The function increases without bound as $x \to \infty$, so there is no absolute minimum at infinity.

### Final Answer:

- The absolute minimum value is $0$ at $x = 7$.
- There is no minimum value at infinity because \( f(x) \to \infty \) as \( x \to \infty \).

</details>

## 10. Find the absolute maximum value on $(0,\infty)$ for $f(x)=14-12x-\dfrac{12}{x}$.

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = 14 - 12x - \frac{12}{x}
$$

We need to find the absolute maximum value of this function on the interval $(0, \infty)$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}\left( 14 - 12x - \frac{12}{x} \right)
$$

Differentiate each term:

$$
f'(x) = -12 + \frac{12}{x^2}
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
-12 + \frac{12}{x^2} = 0
$$

Solve for $x$:

$$
\frac{12}{x^2} = 12
$$

$$
x^2 = 1
$$

$$
x = 1
$$

Since we are working in the interval $(0, \infty)$, we keep $x = 1$.

### Step 4: Evaluate the Function at the Critical Point

We will evaluate the function $f(x)$ at the critical point $x = 1$.

At $x = 1$:

$$
f(1) = 14 - 12(1) - \frac{12}{1} = 14 - 12 - 12 = -10
$$

### Step 5: Analyze Behavior at Infinity and Near Zero

- As $x \to \infty$, the term $-12x$ dominates the behavior, so:

$$
f(x) \to -\infty
$$

- As $x \to 0^+$, the term $-\frac{12}{x}$ dominates, and $f(x) \to -\infty$ as $x \to 0^+$.

### Step 6: Determine the Absolute Maximum

We can summarize the calculations as follows:

| $x$          | Work                                    | $f(x)$  |
|--------------|-----------------------------------------|---------|
| $1$          | $f(1) = 14 - 12(1) - \frac{12}{1}$      | $-10$   |
| $x \to \infty$ | Function tends to negative infinity    | $-\infty$|
| $x \to 0^+$    | Function tends to negative infinity    | $-\infty$|

- The **absolute maximum** value is $-10$ at $x = 1$.

### Final Answer:

- The absolute maximum value is $-10$ at $x = 1$.


</details>

## 11. Find the absolute minimum value on $(0,\infty)$ for $f(x)=x-\dfrac{1}{x}+\dfrac{10}{x^3}$.

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x - \frac{1}{x} + \frac{10}{x^3}
$$

We need to find the absolute minimum value of this function on the interval $(0, \infty)$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}\left( x - \frac{1}{x} + \frac{10}{x^3} \right)
$$

Differentiate each term:

$$
f'(x) = 1 + \frac{1}{x^2} - \frac{30}{x^4}
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
1 + \frac{1}{x^2} - \frac{30}{x^4} = 0
$$

Multiply through by $x^4$ to eliminate the denominators:

$$
x^4 + x^2 - 30 = 0
$$

Let $u = x^2$. This turns the equation into a quadratic:

$$
u^2 + u - 30 = 0
$$

Solve this quadratic equation using the quadratic formula:

$$
u = \frac{-1 \pm \sqrt{1^2 - 4(1)(-30)}}{2(1)}
$$

$$
u = \frac{-1 \pm \sqrt{1 + 120}}{2}
$$

$$
u = \frac{-1 \pm \sqrt{121}}{2}
$$

$$
u = \frac{-1 \pm 11}{2}
$$

Thus, $u = 5$ or $u = -6$. Since $u = x^2$, and $x^2$ cannot be negative, we discard $u = -6$ and keep $u = 5$.

Now, solve for $x$:

$$
x^2 = 5 \quad \Rightarrow \quad x = \sqrt{5}
$$

### Step 4: Evaluate the Function at the Critical Point

We will evaluate the function $f(x)$ at the critical point $x = \sqrt{5}$.

At $x = \sqrt{5}$:

$$
f(\sqrt{5}) = \sqrt{5} - \frac{1}{\sqrt{5}} + \frac{10}{(\sqrt{5})^3}
$$

Simplify each term:

$$
f(\sqrt{5}) = \sqrt{5} - \frac{1}{\sqrt{5}} + \frac{10}{5\sqrt{5}}
$$

$$
f(\sqrt{5}) = \sqrt{5} - \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}}
$$

Combine like terms:

$$
f(\sqrt{5}) = \sqrt{5} + \frac{1}{\sqrt{5}} = \frac{6}{\sqrt{5}}
$$

### Step 5: Analyze Behavior at Infinity and Near Zero

- As $x \to \infty$, the dominant term is $x$, so:

$$
f(x) \to \infty
$$

- As $x \to 0^+$, the terms $-\frac{1}{x}$ and $\frac{10}{x^3}$ dominate, and $f(x) \to \infty$ as $x \to 0^+$.

### Step 6: Determine the Absolute Minimum

We can summarize the calculations as follows:

| $x$          | Work                                    | $f(x)$                     |
|--------------|-----------------------------------------|----------------------------|
| $\sqrt{5}$   | $f(\sqrt{5}) = \frac{6}{\sqrt{5}}$      | $\frac{6}{\sqrt{5}}$        |
| $x \to \infty$ | Function tends to infinity             | $\infty$                   |
| $x \to 0^+$    | Function tends to infinity             | $\infty$                   |

- The **absolute minimum** value is $f(\sqrt{5}) = \frac{6}{\sqrt{5}}$ at $x = \sqrt{5}$.

### Final Answer:

- The absolute minimum value is $\frac{6}{\sqrt{5}}$ at $x = \sqrt{5}$.


</details>

## 12. Find the absolute maximum and minimum, if either exists, for the function on the indicated interval. $f(x)=x^3-6x^2+9x+9$


### a. $[-1,5]$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x^3 - 6x^2 + 9x + 9
$$

We need to find the absolute maximum and minimum values of this function on the interval $[-1, 5]$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}\left( x^3 - 6x^2 + 9x + 9 \right)
$$

Differentiate each term:

$$
f'(x) = 3x^2 - 12x + 9
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
3x^2 - 12x + 9 = 0
$$

Divide through by 3:

$$
x^2 - 4x + 3 = 0
$$

Factor the quadratic:

$$
(x - 3)(x - 1) = 0
$$

Thus, the critical points are:

$$
x = 3 \quad \text{and} \quad x = 1
$$

### Step 4: Evaluate the Function at the Critical Points and Endpoints

We will evaluate the function $f(x)$ at the critical points $x = 1$ and $x = 3$, as well as the endpoints of the interval, $x = -1$ and $x = 5$.

- At $x = 1$:

$$
f(1) = (1)^3 - 6(1)^2 + 9(1) + 9 = 1 - 6 + 9 + 9 = 13
$$

- At $x = 3$:

$$
f(3) = (3)^3 - 6(3)^2 + 9(3) + 9 = 27 - 54 + 27 + 9 = 9
$$

- At $x = -1$ (left endpoint):

$$
f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) + 9 = -1 - 6 - 9 + 9 = -7
$$

- At $x = 5$ (right endpoint):

$$
f(5) = (5)^3 - 6(5)^2 + 9(5) + 9 = 125 - 150 + 45 + 9 = 29
$$

### Step 5: Analyze Behavior at Critical Points and Endpoints

We can summarize the function values at the critical points and endpoints as follows:

| $x$          | Work                                      | $f(x)$  |
|--------------|-------------------------------------------|---------|
| $-1$         | $f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) + 9$    | $-7$    |
| $1$          | $f(1) = (1)^3 - 6(1)^2 + 9(1) + 9$        | $13$    |
| $3$          | $f(3) = (3)^3 - 6(3)^2 + 9(3) + 9$        | $9$     |
| $5$          | $f(5) = (5)^3 - 6(5)^2 + 9(5) + 9$        | $29$    |

### Step 6: Determine the Absolute Maximum and Minimum

- The **absolute minimum** value is $-7$ at $x = -1$.
- The **absolute maximum** value is $29$ at $x = 5$.

### Final Answer:

- The absolute minimum value is $-7$ at $x = -1$.
- The absolute maximum value is $29$ at $x = 5$.


</details>


### b. $[-1,3]$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x^3 - 6x^2 + 9x + 9
$$

We need to find the absolute maximum and minimum values of this function on the interval $[-1, 3]$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}\left( x^3 - 6x^2 + 9x + 9 \right)
$$

Differentiate each term:

$$
f'(x) = 3x^2 - 12x + 9
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
3x^2 - 12x + 9 = 0
$$

Divide through by 3:

$$
x^2 - 4x + 3 = 0
$$

Factor the quadratic:

$$
(x - 3)(x - 1) = 0
$$

Thus, the critical points are:

$$
x = 3 \quad \text{and} \quad x = 1
$$

Since $x = 3$ is included in the interval $[-1, 3]$, we will evaluate the function at both $x = 1$ and $x = 3$.

### Step 4: Evaluate the Function at the Critical Points and Endpoints

We will evaluate the function $f(x)$ at the critical points $x = 1$ and $x = 3$, as well as the endpoints of the interval, $x = -1$ and $x = 3$.

- At $x = 1$:

$$
f(1) = (1)^3 - 6(1)^2 + 9(1) + 9 = 1 - 6 + 9 + 9 = 13
$$

- At $x = 3$:

$$
f(3) = (3)^3 - 6(3)^2 + 9(3) + 9 = 27 - 54 + 27 + 9 = 9
$$

- At $x = -1$ (left endpoint):

$$
f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) + 9 = -1 - 6 - 9 + 9 = -7
$$

### Step 5: Analyze Behavior at Critical Points and Endpoints

We can summarize the function values at the critical points and endpoints as follows:

| $x$          | Work                                      | $f(x)$  |
|--------------|-------------------------------------------|---------|
| $-1$         | $f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) + 9$    | $-7$    |
| $1$          | $f(1) = (1)^3 - 6(1)^2 + 9(1) + 9$        | $13$    |
| $3$          | $f(3) = (3)^3 - 6(3)^2 + 9(3) + 9$        | $9$     |

### Step 6: Determine the Absolute Maximum and Minimum

- The **absolute minimum** value is $-7$ at $x = -1$.
- The **absolute maximum** value is $13$ at $x = 1$.

### Final Answer:

- The absolute minimum value is $-7$ at $x = -1$.
- The absolute maximum value is $13$ at $x = 1$.


</details>


### c. $[2,5]$

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

### Step 1: Function and Interval

We are given the function:

$$
f(x) = x^3 - 6x^2 + 9x + 9
$$

We need to find the absolute maximum and minimum values of this function on the interval $[2, 5]$.

### Step 2: Take the Derivative

To find critical points, we first take the derivative of $f(x)$:

$$
f'(x) = \frac{d}{dx}\left( x^3 - 6x^2 + 9x + 9 \right)
$$

Differentiate each term:

$$
f'(x) = 3x^2 - 12x + 9
$$

### Step 3: Set the Derivative Equal to Zero

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

$$
3x^2 - 12x + 9 = 0
$$

Divide through by 3:

$$
x^2 - 4x + 3 = 0
$$

Factor the quadratic:

$$
(x - 3)(x - 1) = 0
$$

Thus, the critical points are:

$$
x = 3 \quad \text{and} \quad x = 1
$$

Since we are only interested in the interval $[2, 5]$, we discard $x = 1$ (as it lies outside the interval) and keep $x = 3$.

### Step 4: Evaluate the Function at the Critical Point and Endpoints

We will evaluate the function $f(x)$ at the critical point $x = 3$ and at the endpoints of the interval, $x = 2$ and $x = 5$.

- At $x = 3$:

$$
f(3) = (3)^3 - 6(3)^2 + 9(3) + 9 = 27 - 54 + 27 + 9 = 9
$$

- At $x = 2$ (left endpoint):

$$
f(2) = (2)^3 - 6(2)^2 + 9(2) + 9 = 8 - 24 + 18 + 9 = 11
$$

- At $x = 5$ (right endpoint):

$$
f(5) = (5)^3 - 6(5)^2 + 9(5) + 9 = 125 - 150 + 45 + 9 = 29
$$

### Step 5: Analyze Behavior at Critical Points and Endpoints

We can summarize the function values at the critical point and endpoints as follows:

| $x$          | Work                                      | $f(x)$  |
|--------------|-------------------------------------------|---------|
| $2$          | $f(2) = (2)^3 - 6(2)^2 + 9(2) + 9$        | $11$    |
| $3$          | $f(3) = (3)^3 - 6(3)^2 + 9(3) + 9$        | $9$     |
| $5$          | $f(5) = (5)^3 - 6(5)^2 + 9(5) + 9$        | $29$    |

### Step 6: Determine the Absolute Maximum and Minimum

- The **absolute minimum** value is $9$ at $x = 3$.
- The **absolute maximum** value is $29$ at $x = 5$.

### Final Answer:

- The absolute minimum value is $9$ at $x = 3$.
- The absolute maximum value is $29$ at $x = 5$.


</details>


## 13. Express the given quantity as a function $f(x)$ of one variable $x$. The product of two numbers x and y whose sum is 18. $f(x)=$

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



We are tasked with finding a function $f(x)$ that represents the product of two numbers, $x$ and $y$, whose sum is 18.

### Step 1: Set Up the Relationship Between $x$ and $y$

We are given that the sum of two numbers is 18:

$$
x + y = 18
$$

From this, we can solve for $y$ in terms of $x$:

$$
y = 18 - x
$$

### Step 2: Express the Product as a Function of $x$

The product of the two numbers, $x$ and $y$, is given by:

$$
P = x \cdot y
$$

Substitute the expression for $y$ in terms of $x$ from Step 1:

$$
P = x \cdot (18 - x)
$$

### Step 3: Define the Function $f(x)$

Thus, the product can be written as a function of $x$:

$$
f(x) = x(18 - x)
$$

Simplify the expression:

$$
f(x) = 18x - x^2
$$

### Final Answer:

The function that represents the product of the two numbers is:

$$
f(x) = 18x - x^2
$$


</details>
