Function Notation Card Match | ioasarris.sites.sch.gr - Free Printable

Let's solve each problem one by one, explaining the steps clearly.

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Problem 11

Given:
$$
f(x) =
\begin{cases}
x^2 + 2, & x < 1 \\
x - 5, & x \geq 1
\end{cases}
$$
Find $ f(-2) $ and $ f(3) $.

Step 1: Evaluate $ f(-2) $

- Since $ -2 < 1 $, use the first case: $ f(x) = x^2 + 2 $
- $ f(-2) = (-2)^2 + 2 = 4 + 2 = 6 $

Step 2: Evaluate $ f(3) $

- Since $ 3 \geq 1 $, use the second case: $ f(x) = x - 5 $
- $ f(3) = 3 - 5 = -2 $

✔ Answer: $ f(-2) = 6 $, $ f(3) = -2 $

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Problem 12

Given:
$$
f(x) =
\begin{cases}
x^2 + x - 3, & x < -3 \\
2x + 1, & x \geq -3
\end{cases}
$$
Find $ f(-2) $ and $ f(3) $.

Step 1: $ f(-2) $

- $ -2 \geq -3 $ → use second case: $ f(x) = 2x + 1 $
- $ f(-2) = 2(-2) + 1 = -4 + 1 = -3 $

Step 2: $ f(3) $

- $ 3 \geq -3 $ → use second case again
- $ f(3) = 2(3) + 1 = 6 + 1 = 7 $

✔ Answer: $ f(-2) = -3 $, $ f(3) = 7 $

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Problem 13

Given:
$$
f(x) =
\begin{cases}
-x^2 + 1, & x < -2 \\
2x - 7, & x \geq -2
\end{cases}
$$
Find $ f(-2) $ and $ f(3) $.

Step 1: $ f(-2) $

- $ -2 \geq -2 $ → use second case: $ f(x) = 2x - 7 $
- $ f(-2) = 2(-2) - 7 = -4 - 7 = -11 $

Step 2: $ f(3) $

- $ 3 \geq -2 $ → same case
- $ f(3) = 2(3) - 7 = 6 - 7 = -1 $

✔ Answer: $ f(-2) = -11 $, $ f(3) = -1 $

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Problem 14

Given:
$$
f(x) = \sqrt{-5x + 6}
$$
Find $ f(-2) $ and $ f(3) $.

Note: The domain requires $ -5x + 6 \geq 0 $ (since square root is only defined for non-negative values).

Step 1: $ f(-2) $

- Plug in: $ f(-2) = \sqrt{-5(-2) + 6} = \sqrt{10 + 6} = \sqrt{16} = 4 $

Step 2: $ f(3) $

- $ f(3) = \sqrt{-5(3) + 6} = \sqrt{-15 + 6} = \sqrt{-9} $ → not real

⚠️ So $ f(3) $ is undefined in real numbers.

✔ Answer: $ f(-2) = 4 $, $ f(3) $ is undefined (or not real)

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Problem 15

Given:
$$
f(x) = \sqrt{-2x + 5}
$$
Find $ f(-2) $ and $ f(3) $.

Domain condition: $ -2x + 5 \geq 0 \Rightarrow x \leq \frac{5}{2} $

Step 1: $ f(-2) $

- $ f(-2) = \sqrt{-2(-2) + 5} = \sqrt{4 + 5} = \sqrt{9} = 3 $

Step 2: $ f(3) $

- $ f(3) = \sqrt{-2(3) + 5} = \sqrt{-6 + 5} = \sqrt{-1} $ → not real

✔ Answer: $ f(-2) = 3 $, $ f(3) $ is undefined (not real)

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Problem 16

Given:
$$
f(x) = \sqrt{2x + 5}
$$
Find $ f(-2) $ and $ f(3) $.

Domain: $ 2x + 5 \geq 0 \Rightarrow x \geq -\frac{5}{2} $

Step 1: $ f(-2) $

- $ -2 \geq -2.5 $ → valid
- $ f(-2) = \sqrt{2(-2) + 5} = \sqrt{-4 + 5} = \sqrt{1} = 1 $

Step 2: $ f(3) $

- $ f(3) = \sqrt{2(3) + 5} = \sqrt{6 + 5} = \sqrt{11} $

✔ Answer: $ f(-2) = 1 $, $ f(3) = \sqrt{11} $

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Problem 17

Given:
$$
f(x) = 2x^2 + x
$$
Find $ f(x+h) $

Replace every $ x $ with $ x+h $:

$$
f(x+h) = 2(x+h)^2 + (x+h)
= 2(x^2 + 2xh + h^2) + x + h
= 2x^2 + 4xh + 2h^2 + x + h
$$

✔ Answer: $ f(x+h) = 2x^2 + 4xh + 2h^2 + x + h $

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Problem 18

Given:
$$
f(x) = 2x^2 - 3x
$$
Find $ f(x+h) $

Replace $ x $ with $ x+h $:

$$
f(x+h) = 2(x+h)^2 - 3(x+h)
= 2(x^2 + 2xh + h^2) - 3x - 3h
= 2x^2 + 4xh + 2h^2 - 3x - 3h
$$

✔ Answer: $ f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h $

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Problem 19

Given:
$$
f(x) = x^2 - 3
$$
Find $ f(x+h) $

$$
f(x+h) = (x+h)^2 - 3 = x^2 + 2xh + h^2 - 3
$$

✔ Answer: $ f(x+h) = x^2 + 2xh + h^2 - 3 $

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Problem 20

Given:
$$
f(x) = x^2 + 3
$$
Find $ f(x+h) $

$$
f(x+h) = (x+h)^2 + 3 = x^2 + 2xh + h^2 + 3
$$

✔ Answer: $ f(x+h) = x^2 + 2xh + h^2 + 3 $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 11 | $ f(-2) = 6 $, $ f(3) = -2 $ |
| 12 | $ f(-2) = -3 $, $ f(3) = 7 $ |
| 13 | $ f(-2) = -11 $, $ f(3) = -1 $ |
| 14 | $ f(-2) = 4 $, $ f(3) $ undefined (complex) |
| 15 | $ f(-2) = 3 $, $ f(3) $ undefined (complex) |
| 16 | $ f(-2) = 1 $, $ f(3) = \sqrt{11} $ |
| 17 | $ f(x+h) = 2x^2 + 4xh + 2h^2 + x + h $ |
| 18 | $ f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h $ |
| 19 | $ f(x+h) = x^2 + 2xh + h^2 - 3 $ |
| 20 | $ f(x+h) = x^2 + 2xh + h^2 + 3 $ |

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