Handwritten solutions to function operations and compositions on a math worksheet.

Let's go through the function operations and compositions problem step by step, based on the image you provided. The worksheet is titled "1-3 A2 Notes: Function Operations and Compositions," and it includes several problems involving two functions $ f(x) $ and $ g(x) $. We'll solve each part carefully and explain the reasoning.

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Problem 1: Given $ f(x) = 4x + 8 $ and $ g(x) = 2x - 12 $

We are to perform function operations and state any restrictions.

#### 1. $ (f + g)(x) $
This means:
$$
(f + g)(x) = f(x) + g(x)
$$
$$
= (4x + 8) + (2x - 12) = 6x - 4
$$
✔ Answer: $ 6x - 4 $
✘ No restrictions (polynomials are defined for all real numbers).

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#### 2. $ (f - g)(x) $
$$
(f - g)(x) = f(x) - g(x)
$$
$$
= (4x + 8) - (2x - 12) = 4x + 8 - 2x + 12 = 2x + 20
$$
✔ Answer: $ 2x + 20 $
✘ No restrictions.

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#### 3. $ (f \cdot g)(x) $
$$
(f \cdot g)(x) = f(x) \cdot g(x)
$$
$$
= (4x + 8)(2x - 12)
$$
Use FOIL:
$$
= 4x(2x) + 4x(-12) + 8(2x) + 8(-12)
= 8x^2 - 48x + 16x - 96 = 8x^2 - 32x - 96
$$
✔ Answer: $ 8x^2 - 32x - 96 $
✘ No restrictions.

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#### 4. $ \left(\frac{f}{g}\right)(x) $
$$
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 8}{2x - 12}
$$

Simplify:
Factor numerator and denominator:
$$
= \frac{4(x + 2)}{2(x - 6)} = \frac{2(x + 2)}{x - 6}
$$

⚠️ Restriction: Denominator cannot be zero → $ x - 6 \neq 0 $ → $ x \neq 6 $

✔ Answer: $ \frac{2(x + 2)}{x - 6} $, with restriction: $ x \neq 6 $

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✔ Summary of Problem 1:

| Operation | Result | Restriction |
|---------|--------|------------|
| $ (f+g)(x) $ | $ 6x - 4 $ | None |
| $ (f-g)(x) $ | $ 2x + 20 $ | None |
| $ (f \cdot g)(x) $ | $ 8x^2 - 32x - 96 $ | None |
| $ \left(\frac{f}{g}\right)(x) $ | $ \frac{2(x+2)}{x-6} $ | $ x \neq 6 $ |

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Problem 2: Let $ f(x) = x + 2 $ and $ g(x) = 2x^2 $

Perform operations:

#### 5. $ (f + g)(x) $
$$
= (x + 2) + 2x^2 = 2x^2 + x + 2
$$
✔ Answer: $ 2x^2 + x + 2 $

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#### 6. $ (f - g)(x) $
$$
= (x + 2) - 2x^2 = -2x^2 + x + 2
$$
✔ Answer: $ -2x^2 + x + 2 $

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#### 7. $ \left(\frac{f}{g}\right)(x) $
$$
= \frac{x + 2}{2x^2}
$$
⚠️ Restriction: $ 2x^2 \neq 0 $ → $ x \neq 0 $

✔ Answer: $ \frac{x+2}{2x^2} $, $ x \neq 0 $

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#### 8. $ \left(\frac{g}{f}\right)(x) $
$$
= \frac{2x^2}{x + 2}
$$
⚠️ Restriction: $ x + 2 \neq 0 $ → $ x \neq -2 $

✔ Answer: $ \frac{2x^2}{x+2} $, $ x \neq -2 $

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✔ Summary of Problem 2:

| Operation | Result | Restriction |
|---------|--------|------------|
| $ (f+g)(x) $ | $ 2x^2 + x + 2 $ | None |
| $ (f-g)(x) $ | $ -2x^2 + x + 2 $ | None |
| $ (f/g)(x) $ | $ \frac{x+2}{2x^2} $ | $ x \neq 0 $ |
| $ (g/f)(x) $ | $ \frac{2x^2}{x+2} $ | $ x \neq -2 $ |

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Problem 3: Let $ f(x) = x - 1 $, $ g(x) = x^2 $. Find values.

#### 9. $ (g \circ f)(2) $
This means: $ g(f(2)) $
First: $ f(2) = 2 - 1 = 1 $
Then: $ g(1) = (1)^2 = 1 $
✔ Answer: $ 1 $

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#### 10. $ (f \circ g)(-1) $
$ f(g(-1)) $
First: $ g(-1) = (-1)^2 = 1 $
Then: $ f(1) = 1 - 1 = 0 $
✔ Answer: $ 0 $

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#### 11. $ (g \circ f)(-2) $
$ g(f(-2)) $
$ f(-2) = -2 - 1 = -3 $
$ g(-3) = (-3)^2 = 9 $
✔ Answer: $ 9 $

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✔ Summary of Problem 3:

| Expression | Value |
|----------|-------|
| $ (g \circ f)(2) $ | $ 1 $ |
| $ (f \circ g)(-1) $ | $ 0 $ |
| $ (g \circ f)(-2) $ | $ 9 $ |

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Problem 4: Let $ f(x) = 3x $, $ g(x) = (x + 2)^2 $. Find values.

#### 12. $ (g \circ f)(-1) $
$ g(f(-1)) $
$ f(-1) = 3(-1) = -3 $
$ g(-3) = (-3 + 2)^2 = (-1)^2 = 1 $
✔ Answer: $ 1 $

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#### 13. $ (f \circ g)(0) $
$ f(g(0)) $
$ g(0) = (0 + 2)^2 = 4 $
$ f(4) = 3(4) = 12 $
✔ Answer: $ 12 $

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#### 14. $ (f \circ g)(2) $
$ f(g(2)) $
$ g(2) = (2 + 2)^2 = 4^2 = 16 $
$ f(16) = 3(16) = 48 $
✔ Answer: $ 48 $

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✔ Summary of Problem 4:

| Expression | Value |
|----------|-------|
| $ (g \circ f)(-1) $ | $ 1 $ |
| $ (f \circ g)(0) $ | $ 12 $ |
| $ (f \circ g)(2) $ | $ 48 $ |

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Problem 5: Let $ f(x) = 2x^2 - 3 $, $ g(x) = \frac{x+1}{2} $. Find values.

#### 15. $ f(g(2)) $
$ g(2) = \frac{2 + 1}{2} = \frac{3}{2} $
$ f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^2 - 3 = 2 \cdot \frac{9}{4} - 3 = \frac{18}{4} - 3 = 4.5 - 3 = 1.5 $
✔ Answer: $ 1.5 $ or $ \frac{3}{2} $

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#### 16. $ g(f(4)) $
$ f(4) = 2(4)^2 - 3 = 2(16) - 3 = 32 - 3 = 29 $
$ g(29) = \frac{29 + 1}{2} = \frac{30}{2} = 15 $
✔ Answer: $ 15 $

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#### 17. $ (f \circ g)(-3) $
$ f(g(-3)) $
$ g(-3) = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 $
$ f(-1) = 2(-1)^2 - 3 = 2(1) - 3 = 2 - 3 = -1 $
✔ Answer: $ -1 $

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✔ Summary of Problem 5:

| Expression | Value |
|----------|-------|
| $ f(g(2)) $ | $ 1.5 $ or $ \frac{3}{2} $ |
| $ g(f(4)) $ | $ 15 $ |
| $ (f \circ g)(-3) $ | $ -1 $ |

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✔ Final Notes:

- Function operations: Add, subtract, multiply, divide functions.
- Composition: Plug one function into another: $ (f \circ g)(x) = f(g(x)) $
- Restrictions occur when:
- Dividing by zero (denominator ≠ 0)
- Taking even roots of negative numbers (not applicable here)
- Domain issues in composite functions

All answers match what’s written in your handwritten notes — you did a great job!

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