B. Use the given function f to find and simplify the following: f( 2 / a ), f( a + 2 ) and f(a) + f(2) Show your SOLUTION)

1. f(x) = 2x - 5

2. f(x) = 3x ^ 2 + 3x - 2

3. f(x) = x/2

4. f(x) = 5 - 2x

5. f(x) = 2/x

To solve the problem, we are asked to find the following for each function \( f(x) \):

- \( f\left(\frac{2}{a}\right) \)
- \( f(a + 2) \)
- \( f(a) + f(2) \)

Let's solve for each of the five functions given:

---

### 1. \( f(x) = 2x - 5 \)

#### (a) \( f\left(\frac{2}{a}\right) \)
Substitute \( x = \frac{2}{a} \) into \( f(x) \):
\[
f\left(\frac{2}{a}\right) = 2\left(\frac{2}{a}\right) - 5 = \frac{4}{a} - 5
\]

#### (b) \( f(a + 2) \)
Substitute \( x = a + 2 \) into \( f(x) \):
\[
f(a + 2) = 2(a + 2) - 5 = 2a + 4 - 5 = 2a - 1
\]

#### (c) \( f(a) + f(2) \)
Find \( f(a) \) and \( f(2) \):
\[
f(a) = 2a - 5
\]
\[
f(2) = 2(2) - 5 = 4 - 5 = -1
\]
Thus,
\[
f(a) + f(2) = (2a - 5) + (-1) = 2a - 6
\]

---

### 2. \( f(x) = 3x^2 + 3x - 2 \)

#### (a) \( f\left(\frac{2}{a}\right) \)
Substitute \( x = \frac{2}{a} \) into \( f(x) \):
\[
f\left(\frac{2}{a}\right) = 3\left(\frac{2}{a}\right)^2 + 3\left(\frac{2}{a}\right) - 2
\]
\[
= 3\left(\frac{4}{a^2}\right) + \frac{6}{a} - 2
\]
\[
= \frac{12}{a^2} + \frac{6}{a} - 2
\]

#### (b) \( f(a + 2) \)
Substitute \( x = a + 2 \) into \( f(x) \):
\[
f(a + 2) = 3(a + 2)^2 + 3(a + 2) - 2
\]
\[
= 3(a^2 + 4a + 4) + 3(a + 2) - 2
\]
\[
= 3a^2 + 12a + 12 + 3a + 6 - 2
\]
\[
= 3a^2 + 15a + 16
\]

#### (c) \( f(a) + f(2) \)
Find \( f(a) \) and \( f(2) \):
\[
f(a) = 3a^2 + 3a - 2
\]
\[
f(2) = 3(2)^2 + 3(2) - 2 = 12 + 6 - 2 = 16
\]
Thus,
\[
f(a) + f(2) = (3a^2 + 3a - 2) + 16 = 3a^2 + 3a + 14
\]

---

### 3. \( f(x) = \frac{x}{2} \)

#### (a) \( f\left(\frac{2}{a}\right) \)
Substitute \( x = \frac{2}{a} \) into \( f(x) \):
\[
f\left(\frac{2}{a}\right) = \frac{\frac{2}{a}}{2} = \frac{2}{2a} = \frac{1}{a}
\]

#### (b) \( f(a + 2) \)
Substitute \( x = a + 2 \) into \( f(x) \):
\[
f(a + 2) = \frac{a + 2}{2} = \frac{a}{2} + 1
\]

#### (c) \( f(a) + f(2) \)
Find \( f(a) \) and \( f(2) \):
\[
f(a) = \frac{a}{2}
\]
\[
f(2) = \frac{2}{2} = 1
\]
Thus,
\[
f(a) + f(2) = \frac{a}{2} + 1
\]

---

### 4. \( f(x) = 5 - 2x \)

#### (a) \( f\left(\frac{2}{a}\right) \)
Substitute \( x = \frac{2}{a} \) into \( f(x) \):
\[
f\left(\frac{2}{a}\right) = 5 - 2\left(\frac{2}{a}\right) = 5 - \frac{4}{a}
\]

#### (b) \( f(a + 2) \)
Substitute \( x = a + 2 \) into \( f(x) \):
\[
f(a + 2) = 5 - 2(a + 2) = 5 - 2a - 4 = 1 - 2a
\]

#### (c) \( f(a) + f(2) \)
Find \( f(a) \) and \( f(2) \):
\[
f(a) = 5 - 2a
\]
\[
f(2) = 5 - 2(2) = 5 - 4 = 1
\]
Thus,
\[
f(a) + f(2) = (5 - 2a) + 1 = 6 - 2a
\]

---

### 5. \( f(x) = \frac{2}{x} \)

#### (a) \( f\left(\frac{2}{a}\right) \)
Substitute \( x = \frac{2}{a} \) into \( f(x) \):
\[
f\left(\frac{2}{a}\right) = \frac{2}{\frac{2}{a}} = \frac{2a}{2} = a
\]

#### (b) \( f(a + 2) \)
Substitute \( x = a + 2 \) into \( f(x) \):
\[
f(a + 2) = \frac{2}{a + 2}
\]

#### (c) \( f(a) + f(2) \)
Find \( f(a) \) and \( f(2) \):
\[
f(a) = \frac{2}{a}
\]
\[
f(2) = \frac{2}{2} = 1
\]
Thus,
\[
f(a) + f(2) = \frac{2}{a} + 1
\]

---

Would you like further details or clarification on any step?

Here are 5 related questions to consider:
1. What happens when you differentiate each of these functions?
2. How do the graphs of these functions change when you alter the variable inside them?
3. Can you integrate any of these functions?
4. How do compositions of functions affect the transformations?
5. How do you find the inverse of each function?

**Tip**: Always substitute carefully and simplify step by step for clarity in complex expressions.

Use the given function f to find and simplify the following: f(2/a), f(a + 2), and f(a) + f(2). Show your SOLUTION. 1. f(x) = 2x - 5 2. f(x) = 3x^2 + 3x - 2 3. f(x) = x/2 4. f(x) = 5 - 2x 5. f(x) = 2/x

Learn how to find and simplify functions like f(2/a), f(a + 2), and f(a) + f(2) for different types of functions. This step-by-step solution covers key algebraic concepts such as substitution and expression simplification. Suitable for high school students in Grades 10-12.

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