Composition of Functions Answers - ID: 1 Amth 002 Practice ... - Free Printable

Problem: Perform the indicated operations on composite functions.

The task involves performing arithmetic operations (addition, subtraction, multiplication, and division) on composite functions. Each part of the problem provides two functions, $ f(x) $ and $ g(x) $, along with their domains, and asks to compute a specific operation involving these functions.

We will solve each part step by step.

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Part 1:

#### Given:
- $ f(x) = x^2 + 2x $
- $ g(x) = \sqrt{x} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 0 $

#### Task: Compute $ (f + g)(x) $.

#### Solution:
The sum of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(f + g)(x) = f(x) + g(x)
$$
Substitute the given functions:
$$
(f + g)(x) = (x^2 + 2x) + \sqrt{x}
$$
Simplify:
$$
(f + g)(x) = x^2 + 2x + \sqrt{x}
$$

#### Domain:
The domain of $ (f + g)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 0 $, the domain of $ (f + g)(x) $ is:
$$
x \geq 0
$$

#### Final Answer for Part 1:
$$
\boxed{(f + g)(x) = x^2 + 2x + \sqrt{x}, \quad x \geq 0}
$$

---

Part 2:

#### Given:
- $ f(x) = x^2 - 4 $
- $ g(x) = -x - 4 $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: All real numbers ($ \mathbb{R} $)

#### Task: Compute $ (f - g)(x) $.

#### Solution:
The difference of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(f - g)(x) = f(x) - g(x)
$$
Substitute the given functions:
$$
(f - g)(x) = (x^2 - 4) - (-x - 4)
$$
Simplify:
$$
(f - g)(x) = x^2 - 4 + x + 4
$$
$$
(f - g)(x) = x^2 + x
$$

#### Domain:
Both $ f(x) $ and $ g(x) $ are defined for all real numbers, so the domain of $ (f - g)(x) $ is:
$$
x \in \mathbb{R}
$$

#### Final Answer for Part 2:
$$
\boxed{(f - g)(x) = x^2 + x, \quad x \in \mathbb{R}}
$$

---

Part 3:

#### Given:
- $ f(x) = x^2 + 4x $
- $ g(x) = \sqrt{x} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 0 $

#### Task: Compute $ (fg)(x) $.

#### Solution:
The product of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(fg)(x) = f(x) \cdot g(x)
$$
Substitute the given functions:
$$
(fg)(x) = (x^2 + 4x) \cdot \sqrt{x}
$$
Factor $ x $ out of $ x^2 + 4x $:
$$
(fg)(x) = x(x + 4) \cdot \sqrt{x}
$$
Combine terms:
$$
(fg)(x) = x^{3/2}(x + 4)
$$

#### Domain:
The domain of $ (fg)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 0 $, the domain of $ (fg)(x) $ is:
$$
x \geq 0
$$

#### Final Answer for Part 3:
$$
\boxed{(fg)(x) = x^{3/2}(x + 4), \quad x \geq 0}
$$

---

Part 4:

#### Given:
- $ f(x) = x^2 - 4 $
- $ g(x) = -x - 4 $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: All real numbers ($ \mathbb{R} $)

#### Task: Compute $ \left(\frac{f}{g}\right)(x) $.

#### Solution:
The quotient of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}
$$
Substitute the given functions:
$$
\left(\frac{f}{g}\right)(x) = \frac{x^2 - 4}{-x - 4}
$$
Factor the numerator $ x^2 - 4 $ as a difference of squares:
$$
x^2 - 4 = (x - 2)(x + 2)
$$
So:
$$
\left(\frac{f}{g}\right)(x) = \frac{(x - 2)(x + 2)}{-x - 4}
$$
Simplify the denominator $ -x - 4 $ as $ -(x + 4) $:
$$
\left(\frac{f}{g}\right)(x) = \frac{(x - 2)(x + 2)}{-(x + 4)}
$$
$$
\left(\frac{f}{g}\right)(x) = -\frac{(x - 2)(x + 2)}{x + 4}
$$

#### Domain:
The domain of $ \left(\frac{f}{g}\right)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $, excluding any values that make the denominator zero. Since $ f(x) $ and $ g(x) $ are defined for all real numbers, the only restriction comes from the denominator $ x + 4 \neq 0 $, which implies:
$$
x \neq -4
$$
Thus, the domain is:
$$
x \in \mathbb{R}, \quad x \neq -4
$$

#### Final Answer for Part 4:
$$
\boxed{\left(\frac{f}{g}\right)(x) = -\frac{(x - 2)(x + 2)}{x + 4}, \quad x \neq -4}
$$

---

Part 5:

#### Given:
- $ f(x) = x + 2 $
- $ g(x) = \sqrt{x - 2} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 2 $

#### Task: Compute $ (f + g)(x) $.

#### Solution:
The sum of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(f + g)(x) = f(x) + g(x)
$$
Substitute the given functions:
$$
(f + g)(x) = (x + 2) + \sqrt{x - 2}
$$
Simplify:
$$
(f + g)(x) = x + 2 + \sqrt{x - 2}
$$

#### Domain:
The domain of $ (f + g)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 2 $, the domain of $ (f + g)(x) $ is:
$$
x \geq 2
$$

#### Final Answer for Part 5:
$$
\boxed{(f + g)(x) = x + 2 + \sqrt{x - 2}, \quad x \geq 2}
$$

---

Part 6:

#### Given:
- $ f(x) = x^2 - 4 $
- $ g(x) = \sqrt{x - 2} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 2 $

#### Task: Compute $ (f - g)(x) $.

#### Solution:
The difference of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(f - g)(x) = f(x) - g(x)
$$
Substitute the given functions:
$$
(f - g)(x) = (x^2 - 4) - \sqrt{x - 2}
$$
Simplify:
$$
(f - g)(x) = x^2 - 4 - \sqrt{x - 2}
$$

#### Domain:
The domain of $ (f - g)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 2 $, the domain of $ (f - g)(x) $ is:
$$
x \geq 2
$$

#### Final Answer for Part 6:
$$
\boxed{(f - g)(x) = x^2 - 4 - \sqrt{x - 2}, \quad x \geq 2}
$$

---

Part 7:

#### Given:
- $ f(x) = x^2 - 4 $
- $ g(x) = \sqrt{x - 2} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 2 $

#### Task: Compute $ (fg)(x) $.

#### Solution:
The product of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
(fg)(x) = f(x) \cdot g(x)
$$
Substitute the given functions:
$$
(fg)(x) = (x^2 - 4) \cdot \sqrt{x - 2}
$$
Factor the numerator $ x^2 - 4 $ as a difference of squares:
$$
x^2 - 4 = (x - 2)(x + 2)
$$
So:
$$
(fg)(x) = (x - 2)(x + 2) \cdot \sqrt{x - 2}
$$
Combine terms:
$$
(fg)(x) = (x - 2)^{3/2}(x + 2)
$$

#### Domain:
The domain of $ (fg)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 2 $, the domain of $ (fg)(x) $ is:
$$
x \geq 2
$$

#### Final Answer for Part 7:
$$
\boxed{(fg)(x) = (x - 2)^{3/2}(x + 2), \quad x \geq 2}
$$

---

Part 8:

#### Given:
- $ f(x) = x^2 - 2x $
- $ g(x) = \sqrt{x - 2} $
- Domain of $ f(x) $: All real numbers ($ \mathbb{R} $)
- Domain of $ g(x) $: $ x \geq 2 $

#### Task: Compute $ \left(\frac{f}{g}\right)(x) $.

#### Solution:
The quotient of two functions $ f(x) $ and $ g(x) $ is defined as:
$$
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}
$$
Substitute the given functions:
$$
\left(\frac{f}{g}\right)(x) = \frac{x^2 - 2x}{\sqrt{x - 2}}
$$
Factor the numerator $ x^2 - 2x $:
$$
x^2 - 2x = x(x - 2)
$$
So:
$$
\left(\frac{f}{g}\right)(x) = \frac{x(x - 2)}{\sqrt{x - 2}}
$$
Simplify the fraction:
$$
\left(\frac{f}{g}\right)(x) = x \cdot \sqrt{x - 2}
$$

#### Domain:
The domain of $ \left(\frac{f}{g}\right)(x) $ is the intersection of the domains of $ f(x) $ and $ g(x) $, excluding any values that make the denominator zero. Since $ f(x) $ is defined for all real numbers and $ g(x) $ is defined for $ x \geq 2 $, the domain of $ \left(\frac{f}{g}\right)(x) $ is:
$$
x \geq 2
$$

#### Final Answer for Part 8:
$$
\boxed{\left(\frac{f}{g}\right)(x) = x \sqrt{x - 2}, \quad x \geq 2}
$$

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Final Answers:

1. $ \boxed{(f + g)(x) = x^2 + 2x + \sqrt{x}, \quad x \geq 0} $
2. $ \boxed{(f - g)(x) = x^2 + x, \quad x \in \mathbb{R}} $
3. $ \boxed{(fg)(x) = x^{3/2}(x + 4), \quad x \geq 0} $
4. $ \boxed{\left(\frac{f}{g}\right)(x) = -\frac{(x - 2)(x + 2)}{x + 4}, \quad x \neq -4} $
5. $ \boxed{(f + g)(x) = x + 2 + \sqrt{x - 2}, \quad x \geq 2} $
6. $ \boxed{(f - g)(x) = x^2 - 4 - \sqrt{x - 2}, \quad x \geq 2} $
7. $ \boxed{(fg)(x) = (x - 2)^{3/2}(x + 2), \quad x \geq 2} $
8. $ \boxed{\left(\frac{f}{g}\right)(x) = x \sqrt{x - 2}, \quad x \geq 2} $