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It looks like you've uploaded a worksheet titled "Homework 18: Inverse Relations & Functions, Finding Inverses" from a math class. The worksheet has two main sections:

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Section I: Determining if a Graph Has an Inverse

You are given three graphs (labeled A, B, C), and asked to determine whether each graph represents a function that has an inverse.

To determine this, we use the Horizontal Line Test:
> A function has an inverse if and only if it passes the horizontal line test — no horizontal line intersects the graph more than once.

Let’s analyze each graph:

#### Graph A
- This is a parabola opening downward, vertex at the top.
- It fails the horizontal line test because a horizontal line can intersect it in two points.
- So, it does not have an inverse unless the domain is restricted.

✔ Answer: Does NOT have an inverse

#### Graph B
- This is a square root function (like $ y = \sqrt{x} $) starting at the origin and increasing.
- It passes the horizontal line test — each horizontal line hits the graph at most once.
- So, it has an inverse.

✔ Answer: Has an inverse

#### Graph C
- This is a parabola opening upward, vertex at bottom.
- Again, it fails the horizontal line test — horizontal lines cross it twice.
- So, it does not have an inverse unless domain is restricted.

✔ Answer: Does NOT have an inverse

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Section II: Determine if Each Function Has an Inverse

For each function, determine if it has an inverse, and if so, find the inverse. If not, state why.

We’ll go through each one.

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#### 1. $ f(x) = \frac{1}{2}x + 4 $

This is a linear function with slope $ \frac{1}{2} $. Linear functions (with non-zero slope) are one-to-one, so they have inverses.

To find the inverse:
1. Replace $ f(x) $ with $ y $:
$ y = \frac{1}{2}x + 4 $
2. Swap $ x $ and $ y $:
$ x = \frac{1}{2}y + 4 $
3. Solve for $ y $:
$ x - 4 = \frac{1}{2}y $
$ y = 2(x - 4) = 2x - 8 $

✔ Inverse: $ f^{-1}(x) = 2x - 8 $

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#### 2. $ f(x) = (x - 1)^2 $

This is a parabola shifted right by 1 unit. It opens upward, so it fails the horizontal line test (e.g., $ f(0) = f(2) = 1 $).

So, it does not have an inverse over its entire domain.

But if we restrict the domain to $ x \geq 1 $, then it would be one-to-one.

As per instructions: State any restrictions on the domain.

✔ Answer: It doesn’t have an inverse unless the domain is restricted (e.g., $ x \geq 1 $).

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

This is a hyperbola. It is one-to-one (passes horizontal line test) over its domain ($ x \neq 0 $).

To find inverse:
1. $ y = \frac{3}{x} $
2. Swap: $ x = \frac{3}{y} $
3. Solve: $ y = \frac{3}{x} $

So the inverse is the same function!

✔ Inverse: $ f^{-1}(x) = \frac{3}{x} $

Note: Domain restriction: $ x \neq 0 $

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#### 4. $ f(x) = \sqrt{x + 2} $

This is a square root function, defined for $ x \geq -2 $. It increases slowly and passes the horizontal line test.

So it has an inverse.

Find inverse:
1. $ y = \sqrt{x + 2} $
2. Swap: $ x = \sqrt{y + 2} $
3. Square both sides: $ x^2 = y + 2 $
4. $ y = x^2 - 2 $

But remember: original function outputs $ y \geq 0 $, so inverse must have domain $ x \geq 0 $

✔ Inverse: $ f^{-1}(x) = x^2 - 2 $, for $ x \geq 0 $

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#### 5. $ f(x) = x^3 + 3x + 4 $

This is a cubic function. Let's check if it's one-to-one.

Take derivative:
$ f'(x) = 3x^2 + 3 = 3(x^2 + 1) > 0 $ for all $ x $

Since derivative is always positive, the function is strictly increasing, hence one-to-one, so it has an inverse.

But finding the inverse algebraically is very difficult (cubic equations don't have simple inverses in general).

✔ Answer: It has an inverse, but it's not easily expressible in elementary functions.

So, just say: Yes, it has an inverse.

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#### 6. $ f(x) = \frac{2x + 3}{x - 1} $

This is a rational function. Check if it’s one-to-one.

Try to find inverse:
1. $ y = \frac{2x + 3}{x - 1} $
2. Swap: $ x = \frac{2y + 3}{y - 1} $
3. Multiply both sides: $ x(y - 1) = 2y + 3 $
4. $ xy - x = 2y + 3 $
5. Bring terms with $ y $ to one side:
$ xy - 2y = x + 3 $
6. $ y(x - 2) = x + 3 $
7. $ y = \frac{x + 3}{x - 2} $

✔ Inverse: $ f^{-1}(x) = \frac{x + 3}{x - 2} $

Domain: $ x \neq 2 $, since denominator can't be zero.

Also, original function has domain $ x \neq 1 $, so inverse range excludes $ x = 1 $, which matches.

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✔ Final Answers Summary:

#### Section I (Graphs):
- A: Does NOT have an inverse
- B: Has an inverse
- C: Does NOT have an inverse

#### Section II (Functions):
1. $ f(x) = \frac{1}{2}x + 4 $ → Has inverse: $ f^{-1}(x) = 2x - 8 $
2. $ f(x) = (x - 1)^2 $ → Doesn't have an inverse (unless domain restricted)
3. $ f(x) = \frac{3}{x} $ → Has inverse: $ f^{-1}(x) = \frac{3}{x} $, $ x \neq 0 $
4. $ f(x) = \sqrt{x + 2} $ → Has inverse: $ f^{-1}(x) = x^2 - 2 $, $ x \geq 0 $
5. $ f(x) = x^3 + 3x + 4 $ → Has an inverse (function is strictly increasing)
6. $ f(x) = \frac{2x + 3}{x - 1} $ → Has inverse: $ f^{-1}(x) = \frac{x + 3}{x - 2} $, $ x \neq 2 $

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