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# 2.2 - Combinations of Functions

## JIT 12. Multiplying Polynomials

## Multiplying Monomials

- Multiply coefficients, add powers of $x$: $x^a \cdot x^b=x^{a+b}$

### Example 1. $(4x)(6x)=24x^2$

### Example 2. $(4x)(-7)=-28x$

### Example 3. $(2x^5)(-3x^3)=-6x^8$

### Example 4 $(-5)(3x)=-15x$

### Example 5. $(-5x)(-4x)=20x^2$

### Example 6. $(-2x^2)(4x^3)=-8x^5$

## Multiplying binomials: also known as **FOIL**.

- $$(a+b)(c+d)=ac+ad+bc+bd$$
- **FOIL**: **F**irst + **O**uter + **I**nner + **L**ast

### Example 7.

\begin{align}
(x+5)(x+2)&=x \cdot x +x \cdot 2+ 5 \cdot x +5 \cdot 2 \\
&=x^2+2x+5x+10 \\
&=x^2+7x+10
\end{align}

### Example 8.

\begin{align}
(2x-3)(x+4)&=2x \cdot x +2x \cdot 4-3 \cdot x -3\cdot 4 \\
&=2x^2+8x-3x-12 \\
&=2x^2+5x-12
\end{align}

### Example 9.

\begin{align}
(x-4)(3x+1)&=x \cdot 3x +x \cdot 1-4 \cdot 3x -4\cdot 1 \\
&=3x^2+x-12x-4 \\
&=3x^2-11x-4
\end{align}

### Example 10.

\begin{align}
(y-2)(2y+3)&=y \cdot 2y+y \cdot 3-2 \cdot 2y -2 \cdot 3 \\
&=2y^2+3y-4y-6 \\
&=2y^2-y-6
\end{align}

## Squares of binomials

$$(x-5)^2=(x-5)(x-5)=x^2-5x-5x+25=x^2-10x+25$$

$$(2x+1)^2=(2x+1)(2x+1)=4x^2+2x+2x+1=4x^2+4x+1$$

$$(3y-4)^2=(3y-4)(3y-4)=9y^2-12y-12y+16=9y^2-24y+16$$

## Other examples of distribution.

$$2x^3(5x^2-7)=10x^5-14x^3$$

$$-5x^2(-3x-2)=15x^3+10x^2$$

## Difference of Two Squares $$(A+B)(A-B)=A^2-B^2$$

### Example. $$(x+3)(x-3)=x^2-9$$

### Example. $$(4+x)(4-x)=16-x^2$$

# JIT 19. Fractions

$$\dfrac{x^a}{x^b}=x^{a-b}$$

- Key fact: $\dfrac{a}{b}=\dfrac{a \cdot c}{b \cdot c}$

- Key fact: $\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c}$

We often need to do the opposite:

- Key fact: $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$

### Example. Split into three fractions:

\begin{align}
\dfrac{4x^3-2x+1}{2x^2}&=\dfrac{4x^3}{2x^2}-\dfrac{2x}{2x^2}+\dfrac{1}{2x^2} \\
&=2x-x^{-1}+\dfrac{1}{2}x^{-2}
\end{align}

### Example. Split into three fractions:

\begin{align}
\dfrac{3x^5-2x+1}{x^3}&=\dfrac{3x^5}{x^3}-\dfrac{2x}{x^3}+\dfrac{1}{x^3} \\
&=3x^2-2x^{-2}+x^{-3}
\end{align}

### Example. Split into three fractions:

\begin{align}
\dfrac{3x^5-4x+1}{2x}&=\dfrac{3x^5}{2x}-\dfrac{4x}{2x}+\dfrac{1}{2x} \\
&=\dfrac{3}{2}x^4-2+\dfrac{1}{2}x^{-1}
\end{align}

# 2.2 Combinations of Functions

- We learn to add, subtract, multiply, and divide functions, just like with numbers.

$$(f+g)(x)=f(x)+g(x)$$

$$(f-g)(x)=f(x)-g(x)$$

$$(fg)(x)=f(x) g(x)$$

$$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$$

## Example. Let $f(x)=2x-6$ and $g(x)=3x^2+5$

---

### 1. Find $(f+g)(3)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $x = 3$ into $f(x)$ and $g(x)$
$f(3) = 2(3) - 6 = 6 - 6 = 0$
$g(3) = 3(3)^2 + 5 = 3(9) + 5 = 27 + 5 = 32$

#### Step 3: Add the results
$(f+g)(3) = f(3) + g(3) = 0 + 32 = 32$

#### Solution:
$(f+g)(3) = 32$

:::

---

### 2. Find $(f-g)(-1)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $x = -1$ into $f(x)$ and $g(x)$
$f(-1) = 2(-1) - 6 = -2 - 6 = -8$
$g(-1) = 3(-1)^2 + 5 = 3(1) + 5 = 3 + 5 = 8$

#### Step 3: Subtract the results
$(f-g)(-1) = f(-1) - g(-1) = -8 - 8 = -16$

#### Solution:
$(f-g)(-1) = -16$

:::

---

### 3. Find $(g-f)(2)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(g-f)(x)$
$(g-f)(x) = g(x) - f(x)$

#### Step 2: Substitute $x = 2$ into $f(x)$ and $g(x)$
$f(2) = 2(2) - 6 = 4 - 6 = -2$
$g(2) = 3(2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17$

#### Step 3: Subtract the results
$(g-f)(2) = g(2) - f(2) = 17 - (-2) = 17 + 2 = 19$

#### Solution:
$(g-f)(2) = 19$

:::

---

### 4. Find $(fg)(2)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $x = 2$ into $f(x)$ and $g(x)$
$f(2) = 2(2) - 6 = 4 - 6 = -2$
$g(2) = 3(2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17$

#### Step 3: Multiply the results
$(fg)(2) = f(2) \cdot g(2) = -2 \cdot 17 = -34$

#### Solution:
$(fg)(2) = -34$

:::

---

### 5. Find $(f/g)(4)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \frac{f(x)}{g(x)}$

#### Step 2: Substitute $x = 4$ into $f(x)$ and $g(x)$
$f(4) = 2(4) - 6 = 8 - 6 = 2$
$g(4) = 3(4)^2 + 5 = 3(16) + 5 = 48 + 5 = 53$

#### Step 3: Divide the results
$(f/g)(4) = \frac{f(4)}{g(4)} = \frac{2}{53}$

#### Solution:
$(f/g)(4) = \frac{2}{53}$
:::

---

### 6. Find $(f+g)(4)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $x = 4$ into $f(x)$ and $g(x)$
$f(4) = 2(4) - 6 = 8 - 6 = 2$
$g(4) = 3(4)^2 + 5 = 3(16) + 5 = 48 + 5 = 53$

#### Step 3: Add the results
$(f+g)(4) = f(4) + g(4) = 2 + 53 = 55$

#### Solution:
$(f+g)(4) = 55$

:::

---

### 7. Find $(f-g)(-2)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $x = -2$ into $f(x)$ and $g(x)$
$f(-2) = 2(-2) - 6 = -4 - 6 = -10$
$g(-2) = 3(-2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17$

#### Step 3: Subtract the results
$(f-g)(-2) = f(-2) - g(-2) = -10 - 17 = -27$

#### Solution:
$(f-g)(-2) = -27$

:::

---

### 8. Find $(g-f)(3)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(g-f)(x)$
$(g-f)(x) = g(x) - f(x)$

#### Step 2: Substitute $x = 3$ into $f(x)$ and $g(x)$
$f(3) = 2(3) - 6 = 6 - 6 = 0$
$g(3) = 3(3)^2 + 5 = 3(9) + 5 = 27 + 5 = 32$

#### Step 3: Subtract the results
$(g-f)(3) = g(3) - f(3) = 32 - 0 = 32$

#### Solution:
$(g-f)(3) = 32$

:::

---

### 9. Find $(fg)(3)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $x = 3$ into $f(x)$ and $g(x)$
$f(3) = 2(3) - 6 = 6 - 6 = 0$
$g(3) = 3(3)^2 + 5 = 3(9) + 5 = 27 + 5 = 32$

#### Step 3: Multiply the results
$(fg)(3) = f(3) \cdot g(3) = 0 \cdot 32 = 0$

#### Solution:
$(fg)(3) = 0$

:::

---

### 10. Find $(f/g)(2)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \frac{f(x)}{g(x)}$

#### Step 2: Substitute $x = 2$ into $f(x)$ and $g(x)$
$f(2) = 2(2) - 6 = 4 - 6 = -2$
$g(2) = 3(2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17$

#### Step 3: Divide the results
$(f/g)(2) = \frac{f(2)}{g(2)} = \frac{-2}{17}$

#### Solution:
$(f/g)(2) = \frac{-2}{17}$

:::

---

### 11. Find $(f+g)(\frac{1}{2})$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $x = \frac{1}{2}$ into $f(x)$ and $g(x)$
$f(\frac{1}{2}) = 2(\frac{1}{2}) - 6 = 1 - 6 = -5$
$g(\frac{1}{2}) = 3(\frac{1}{2})^2 + 5 = 3(\frac{1}{4}) + 5 = \frac{3}{4} + 5 = \frac{3}{4} + \frac{20}{4} = \frac{23}{4}$

#### Step 3: Add the results
$(f+g)(\frac{1}{2}) = f(\frac{1}{2}) + g(\frac{1}{2}) = -5 + \frac{23}{4} = \frac{-20}{4} + \frac{23}{4} = \frac{3}{4}$

#### Solution:
$(f+g)(\frac{1}{2}) = \frac{3}{4}$

:::

---

### 12. Find $(g-f)(-\frac{1}{2})$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(g-f)(x)$
$(g-f)(x) = g(x) - f(x)$

#### Step 2: Substitute $x = -\frac{1}{2}$ into $f(x)$ and $g(x)$
$f(-\frac{1}{2}) = 2(-\frac{1}{2}) - 6 = -1 - 6 = -7$
$g(-\frac{1}{2}) = 3(-\frac{1}{2})^2 + 5 = 3(\frac{1}{4}) + 5 = \frac{3}{4} + 5 = \frac{3}{4} + \frac{20}{4} = \frac{23}{4}$

#### Step 3: Subtract the results
$(g-f)(-\frac{1}{2}) = g(-\frac{1}{2}) - f(-\frac{1}{2}) = \frac{23}{4} - (-7) = \frac{23}{4} + \frac{28}{4} = \frac{51}{4}$

#### Solution:
$(g-f)(-\frac{1}{2}) = \frac{51}{4}$

:::

---

### 13. Find $(fg)(\frac{1}{3})$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $x = \frac{1}{3}$ into $f(x)$ and $g(x)$
$f(\frac{1}{3}) = 2(\frac{1}{3}) - 6 = \frac{2}{3} - 6 = \frac{2}{3} - \frac{18}{3} = \frac{-16}{3}$
$g(\frac{1}{3}) = 3(\frac{1}{3})^2 + 5 = 3(\frac{1}{9}) + 5 = \frac{3}{9} + 5 = \frac{1}{3} + 5 = \frac{1}{3} + \frac{15}{3} = \frac{16}{3}$

#### Step 3: Multiply the results
$(fg)(\frac{1}{3}) = f(\frac{1}{3}) \cdot g(\frac{1}{3}) = \frac{-16}{3} \cdot \frac{16}{3} = \frac{-256}{9}$

#### Solution:
$(fg)(\frac{1}{3}) = \frac{-256}{9}$

:::

---

### 14. Find $(f/g)(\frac{1}{2})$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \frac{f(x)}{g(x)}$

#### Step 2: Substitute $x = \frac{1}{2}$ into $f(x)$ and $g(x)$
$f(\frac{1}{2}) = 2(\frac{1}{2}) - 6 = 1 - 6 = -5$
$g(\frac{1}{2}) = 3(\frac{1}{2})^2 + 5 = 3(\frac{1}{4}) + 5 = \frac{3}{4} + 5 = \frac{3}{4}+\frac{20}{4}=\frac{23}{4}$

#### Step 3: Divide the results
$(f/g)(\frac{1}{2}) = \frac{f(\frac{1}{2})}{g(\frac{1}{2})} = \frac{-5}{\frac{23}{4}} = \frac{-5 \cdot 4}{23} = \frac{-20}{23}$

#### Solution:
$(f/g)(\frac{1}{2}) = \frac{-20}{23}$

:::

---

## Let $f(x)=2x+5$, $g(x)=x+4$

### 15. Find $(f+g)(x)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $f(x) = 2x + 5$ and $g(x) = x + 4$
$(f+g)(x) = (2x + 5) + (x + 4)$

#### Step 3: Simplify the expression
$(f+g)(x) = 2x + x + 5 + 4 = 3x + 9$

#### Solution:
$(f+g)(x) = 3x + 9$

:::

---

### 16. Find $(f-g)(x)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $f(x) = 2x + 5$ and $g(x) = x + 4$
$(f-g)(x) = (2x + 5) - (x + 4)$
$(f-g)(x) = 2x + 5 -x-4 \qquad$ Distribute the negative.

#### Step 3: Simplify the expression
$(f-g)(x) = 2x - x + 5 - 4 = x + 1$

#### Solution:
$(f-g)(x) = x + 1$

:::

---

### 17. Find $(g-f)(x)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(g-f)(x)$
$(g-f)(x) = g(x) - f(x)$

#### Step 2: Substitute $f(x) = 2x + 5$ and $g(x) = x + 4$
$(g-f)(x) = (x + 4) - (2x + 5)$
$(g-f)(x) = x + 4 - 2x - 5 \qquad$ Distribute the negative.

#### Step 3: Simplify the expression
$(g-f)(x) = x - 2x + 4 - 5 = -x - 1$

#### Solution:
$(g-f)(x) = -x - 1$

:::

---

### 18. Find $(fg)(x)$.

::: spoiler

<summary> Solution: </summary>

We are given two functions:

$f(x) = 2x + 5$
$g(x) = x + 4$

### Calculating $(fg)(x)$ Using the FOIL Method

1. We are asked to find $(fg)(x)$.
2. This means we need to multiply $f(x)$ and $g(x)$ together using the FOIL method (First, Outer, Inner, Last).

- Write the product:
$(fg)(x) = (2x + 5)(x + 4)$

Now, apply FOIL:

- **First** terms: $2x \cdot x = 2x^2$
- **Outer** terms: $2x \cdot 4 = 8x$
- **Inner** terms: $5 \cdot x = 5x$
- **Last** terms: $5 \cdot 4 = 20$

Now, combine all the terms:

$2x^2 + 8x + 5x + 20$

Combine like terms:

$= 2x^2 + 13x + 20$

So, $(fg)(x) = 2x^2 + 13x + 20$.

:::

---

### 19. Find $(f/g)(x)$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \dfrac{f(x)}{g(x)}$

#### Step 2: Substitute $f(x) = 2x + 5$ and $g(x) = x + 4$
$(f/g)(x) = \dfrac{2x + 5}{x + 4}$

#### Solution:
$(f/g)(x) = \dfrac{2x + 5}{x + 4}$

:::

---

### 20. Find $(f+g)(x)$ for $f(x) = 2x - 3$ and $g(x) = x + 5$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $f(x) = 2x - 3$ and $g(x) = x + 5$
$(f+g)(x) = (2x - 3) + (x + 5)=2x-3+x+5$

#### Step 3: Simplify the expression
$(f+g)(x) = 2x + x - 3 + 5 = 3x + 2$

#### Solution:
$(f+g)(x) = 3x + 2$

:::

---

### 21. Find $(f-g)(x)$ for $f(x) = 2x - 3$ and $g(x) = x + 5$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $f(x) = 2x - 3$ and $g(x) = x + 5$
$(f-g)(x) = (2x - 3) - (x + 5)$
$(f-g)(x) = 2x - 3 - x - 5\qquad$ Distribute the negative.

#### Step 3: Simplify the expression
$(f-g)(x) = 2x - x - 3 - 5 = x - 8$

#### Solution:
$(f-g)(x) = x - 8$

:::

---

### 22. Find $(g-f)(x)$ for $f(x) = 2x - 3$ and $g(x) = x + 5$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(g-f)(x)$
$(g-f)(x) = g(x) - f(x)$

#### Step 2: Substitute $f(x) = 2x - 3$ and $g(x) = x + 5$
$(g-f)(x) = (x + 5) - (2x - 3)$
$(g-f)(x) = x + 5 - 2x +3\qquad$ Distribute the negative.

#### Step 3: Simplify the expression
$(g-f)(x) = x - 2x + 5 + 3 = -x + 8$

#### Solution:
$(g-f)(x) = -x + 8$

:::

---

### 23. Find $(fg)(x)$ for $f(x) = 2x - 3$ and $g(x) = x + 5$.

::: spoiler

<summary> Solution: </summary>

We are given two functions:

$f(x) = 2x - 3$
$g(x) = x + 5$

### Calculating $(fg)(x)$ Using the FOIL Method

1. We are asked to find $(fg)(x)$.
2. This means we need to multiply $f(x)$ and $g(x)$ together using the FOIL method (First, Outer, Inner, Last).

- Write the product:
$(fg)(x) = (2x - 3)(x + 5)$

Now, apply FOIL:

- **First** terms: $2x \cdot x = 2x^2$
- **Outer** terms: $2x \cdot 5 = 10x$
- **Inner** terms: $-3 \cdot x = -3x$
- **Last** terms: $-3 \cdot 5 = -15$

Now, combine all the terms:

$2x^2 + 10x - 3x - 15$

Combine like terms:

$= 2x^2 + 7x - 15$

So, $(fg)(x) = 2x^2 + 7x - 15$.

:::

---

### 24. Find $(f/g)(x)$ for $f(x) = 2x - 3$ and $g(x) = x + 5$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \dfrac{f(x)}{g(x)}$

#### Step 2: Substitute $f(x) = 2x - 3$ and $g(x) = x + 5$
$(f/g)(x) = \dfrac{2x - 3}{x + 5}$

#### Solution:
$(f/g)(x) = \dfrac{2x - 3}{x + 5}$

:::

---

### 25. Find $(f+g)(x)$ for $f(x) = 2x^2$ and $g(x) = x - 6$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $f(x) = 2x^2$ and $g(x) = x - 6$
$(f+g)(x) = (2x^2) + (x - 6)$

#### Step 3: Simplify the expression
$(f+g)(x) = 2x^2 + x - 6$

#### Solution:
$(f+g)(x) = 2x^2 + x - 6$

:::

---

### 26. Find $(f-g)(x)$ for $f(x) = 2x^2$ and $g(x) = x - 6$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $f(x) = 2x^2$ and $g(x) = x - 6$
$(f-g)(x) = (2x^2) - (x - 6)$

#### Step 3: Simplify the expression
$(f-g)(x) = 2x^2 - x + 6$

#### Solution:
$(f-g)(x) = 2x^2 - x + 6$

:::

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### 27. Find $(fg)(x)$ for $f(x) = 2x^2$ and $g(x) = x - 6$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $f(x) = 2x^2$ and $g(x) = x - 6$
$(fg)(x) = (2x^2)(x - 6)$

#### Step 3: Expand the expression
$(fg)(x) = 2x^2(x - 6) = 2x^3 - 12x^2$

#### Solution:
$(fg)(x) = 2x^3 - 12x^2$

:::

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### 28. Find $(f+g)(x)$ for $f(x) = 2x^3 + x^2 - 3x - 7$ and $g(x) = x^3 + 7x - 6$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f+g)(x)$
$(f+g)(x) = f(x) + g(x)$

#### Step 2: Substitute $f(x) = 2x^3 + x^2 - 3x - 7$ and $g(x) = x^3 + 7x - 6$
$(f+g)(x) = (2x^3 + x^2 - 3x - 7) + (x^3 + 7x - 6)$

#### Step 3: Combine like terms
- Combine the $x^3$ terms: $2x^3 + x^3 = 3x^3$
- Combine the $x^2$ terms: $x^2$
- Combine the $x$ terms: $-3x + 7x = 4x$
- Combine the constant terms: $-7 - 6 = -13$

#### Step 4: Simplified expression
$(f+g)(x) = 3x^3 + x^2 + 4x - 13$

#### Solution:
$(f+g)(x) = 3x^3 + x^2 + 4x - 13$

:::

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### 29. Find $(f-g)(x)$ for $f(x) = 2x^3 + x^2 - 3x - 7$ and $g(x) = x^3 + 7x - 6$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f-g)(x)$
$(f-g)(x) = f(x) - g(x)$

#### Step 2: Substitute $f(x) = 2x^3 + x^2 - 3x - 7$ and $g(x) = x^3 + 7x - 6$
$(f-g)(x) = (2x^3 + x^2 - 3x - 7) - (x^3 + 7x - 6)$

#### Step 3: Distribute the negative sign
$(f-g)(x) = 2x^3 + x^2 - 3x - 7 - x^3 - 7x + 6$

#### Step 4: Combine like terms
- Combine the $x^3$ terms: $2x^3 - x^3 = x^3$
- Combine the $x^2$ terms: $x^2$
- Combine the $x$ terms: $-3x - 7x = -10x$
- Combine the constant terms: $-7 + 6 = -1$

#### Step 5: Simplified expression
$(f-g)(x) = x^3 + x^2 - 10x - 1$

#### Solution:
$(f-g)(x) = x^3 + x^2 - 10x - 1$

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### 30. Find $(fg)(x)$ for $f(x) = 2x^5 + x - 1$ and $g(x) = x^2$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $f(x) = 2x^5 + x - 1$ and $g(x) = x^2$
$(fg)(x) = (2x^5 + x - 1) \cdot x^2$

#### Step 3: Distribute $x^2$ across all terms
$(fg)(x) = 2x^5 \cdot x^2 + x \cdot x^2 - 1 \cdot x^2$

#### Step 4: Simplify each term
- $2x^5 \cdot x^2 = 2x^7$
- $x \cdot x^2 = x^3$
- $-1 \cdot x^2 = -x^2$

#### Step 5: Simplified expression
$(fg)(x) = 2x^7 + x^3 - x^2$

#### Solution:
$(fg)(x) = 2x^7 + x^3 - x^2$

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### 31. Find $(f/g)(x)$ for $f(x) = 2x^5 + x - 1$ and $g(x) = x^2$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(f/g)(x)$
$(f/g)(x) = \dfrac{f(x)}{g(x)}$

#### Step 2: Substitute $f(x) = 2x^5 + x - 1$ and $g(x) = x^2$
$(f/g)(x) = \dfrac{2x^5 + x - 1}{x^2}$

#### Step 3: Break up the fraction
$(f/g)(x) = \dfrac{2x^5}{x^2} + \dfrac{x}{x^2} - \dfrac{1}{x^2}$

#### Step 4: Simplify each term
- $\dfrac{2x^5}{x^2} = 2x^3$
- $\dfrac{x}{x^2} = x^{-1}$
- $\dfrac{1}{x^2}=x^{-2}$

#### Step 5: Simplified expression
$(f/g)(x) = 2x^3 + x^{-1} - x^{-2}$

#### Solution:
$(f/g)(x) = 2x^3 + x^{-1} - x^{-2}$

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### 32. Find $(fg)(4)$ for $f(x) = x + 4$ and $g(x) = \sqrt{x - 3}$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $x = 4$ into $f(x)$ and $g(x)$
$f(4) = 4 + 4 = 8$
$g(4) = \sqrt{4 - 3} = \sqrt{1} = 1$

#### Step 3: Multiply the results
$(fg)(4) = f(4) \cdot g(4) = 8 \cdot 1 = 8$

#### Solution:
$(fg)(4) = 8$

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### 33. Find $(fg)(1)$ for $f(x) = x + 4$ and $g(x) = \sqrt{x - 3}$.

::: spoiler

<summary> Solution: </summary>

#### Step 1: Definition of $(fg)(x)$
$(fg)(x) = f(x) \cdot g(x)$

#### Step 2: Substitute $x = 1$ into $f(x)$ and $g(x)$
$f(1) = 1 + 4 = 5$
$g(1) = \sqrt{1 - 3} = \sqrt{-2}$ (undefined for real numbers)

#### Solution:
$(fg)(1)$ is undefined because $g(1) = \sqrt{-2}$ is not a real number.

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## Cost, Revenue, and Profit

Let $x$ be the number produced of some product in production.
- $R(x)$ is the **revenue** obtained by selling x units of the product. That is:
the amount of money buyers pay for x units of the product.
- $C(x)$ is the **cost** of producing x units. This includes labor, raw materials,
facilities, etc.
- $P(x)$ is the **profit** obtained by the production and sale of x units. That
is, the revenue minus the cost of production:

$$P(x)=R(x)-C(x)$$

### Example. The cost of producing $x$ cell phones is $C(x)=1000+200x$. They sell for $800 each. What is the profit from selling $x$ phones?

- If they sell for $800 each, the revenue is $R(x)=800x$. Then:

\begin{align}
P(x)&=R(x)-C(x) \\
&=800x-(1000+200x) \\
&=800x-1000-200x \qquad Distribute the negative.\\
&=600x-1000
\end{align}

- Note that $C(x)$ can be seen as the sum of fixed costs ($1000) and variable costs ($2 per phone.)

### Example. The cost of producing $x$ candles is $C(x)=100+3x$. They sell for $5 each. What is the profit from selling $x$ phones?

- The revenue is $R(x)=5x$

\begin{align}
P(x)&=R(x)-C(x) \\
&=5x-(100+3x) \\
&=5x-100-3x \\
&=2x-100
\end{align}