D. FREE RESPONSE. Show complete solution. (10 points)
1. Find f(g(x)) when f(x) = 3x2 + 2 and g(x) = √1 − x
2. Given the function f(x) = 5x2 – 3x + 2 and g(x) = x + 3, find f(g(1)) and g(f(1)).
3. If f and g are two distinct functions, is f(g(x)) = g(f(x))? Justify your answer.

### 1. Find \( f(g(x)) \) when \( f(x) = 3x^2 + 2 \) and \( g(x) = \sqrt{1 - x} \)

#### Step 1: Substitute \( g(x) = \sqrt{1 - x} \) into \( f(x) \)
We are asked to find \( f(g(x)) \), which means we need to plug the expression for \( g(x) \) into the function \( f(x) \).

The function \( f(x) \) is:
\[
f(x) = 3x^2 + 2
\]
Now, substitute \( g(x) = \sqrt{1 - x} \) into \( f(x) \):
\[
f(g(x)) = 3(\sqrt{1 - x})^2 + 2
\]

#### Step 2: Simplify
Since \( (\sqrt{1 - x})^2 = 1 - x \), we have:
\[
f(g(x)) = 3(1 - x) + 2
\]
Simplify further:
\[
f(g(x)) = 3 - 3x + 2
\]
\[
f(g(x)) = 5 - 3x
\]

Thus, \( f(g(x)) = 5 - 3x \).

---

### 2. Find \( f(g(1)) \) and \( g(f(1)) \) given \( f(x) = 5x^2 - 3x + 2 \) and \( g(x) = x + 3 \)

#### Part (a) Find \( f(g(1)) \)

##### Step 1: Compute \( g(1) \)
We need to first find \( g(1) \). Since \( g(x) = x + 3 \), substitute \( x = 1 \):
\[
g(1) = 1 + 3 = 4
\]

##### Step 2: Compute \( f(g(1)) = f(4) \)
Now we substitute \( g(1) = 4 \) into \( f(x) \). The function \( f(x) = 5x^2 - 3x + 2 \), so:
\[
f(4) = 5(4)^2 - 3(4) + 2
\]
Simplify:
\[
f(4) = 5(16) - 12 + 2 = 80 - 12 + 2 = 70
\]

Thus, \( f(g(1)) = 70 \).

#### Part (b) Find \( g(f(1)) \)

##### Step 1: Compute \( f(1) \)
We need to first find \( f(1) \). Since \( f(x) = 5x^2 - 3x + 2 \), substitute \( x = 1 \):
\[
f(1) = 5(1)^2 - 3(1) + 2 = 5 - 3 + 2 = 4
\]

##### Step 2: Compute \( g(f(1)) = g(4) \)
Now substitute \( f(1) = 4 \) into \( g(x) \). Since \( g(x) = x + 3 \), we have:
\[
g(4) = 4 + 3 = 7
\]

Thus, \( g(f(1)) = 7 \).

---

### 3. Is \( f(g(x)) = g(f(x)) \) for two distinct functions?

In general, \( f(g(x)) \) is **not** equal to \( g(f(x)) \) for two distinct functions. This is because composition of functions is generally **not commutative**.

To justify this, let’s look at an example with two distinct functions:
- Let \( f(x) = 2x \) and \( g(x) = x + 1 \).

Now, compute both compositions:
- \( f(g(x)) = 2(g(x)) = 2(x + 1) = 2x + 2 \),
- \( g(f(x)) = g(2x) = 2x + 1 \).

Clearly, \( f(g(x)) = 2x + 2 \) and \( g(f(x)) = 2x + 1 \), so \( f(g(x)) \neq g(f(x)) \).

Thus, \( f(g(x)) \neq g(f(x)) \) in general.

---

### Summary of Results:
1. \( f(g(x)) = 5 - 3x \)
2. \( f(g(1)) = 70 \) and \( g(f(1)) = 7 \)
3. In general, \( f(g(x)) \neq g(f(x)) \), because function composition is not commutative.

---

Do you want further explanations or have any other questions?

### Related Questions:
1. How do you find \( g(f(g(x))) \) for given functions?
2. What is the domain of \( f(g(x)) \) when \( f(x) = \frac{1}{x} \) and \( g(x) = \sqrt{x} \)?
3. What is the general property of function composition?
4. If \( f(x) = x^2 \) and \( g(x) = x + 2 \), what is \( g(g(x)) \)?
5. Is \( f(f(x)) \) always equal to \( f(x)^2 \)?

### Tip:
Function composition is not commutative. Always check the order in which functions are composed.

1. Find f(g(x)) when f(x) = 3x^2 + 2 and g(x) = √1 − x
2. Given the function f(x) = 5x^2 – 3x + 2 and g(x) = x + 3, find f(g(1)) and g(f(1)).
3. If f and g are two distinct functions, is f(g(x)) = g(f(x))? Justify your answer.

Learn how to solve function composition problems, including finding f(g(x)) when f(x) = 3x^2 + 2 and g(x) = √1 − x. This guide also explores the non-commutative nature of function composition with detailed step-by-step solutions. Suitable for Grades 10-12.

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