Math practice problems focusing on algebraic simplification and equation solving.

Let’s solve each part step by step.

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Problem 1: Find the inverse of each function.

We are given functions and asked to find their inverses. To find the inverse of a function, we:

- Replace f(x) with y
- Swap x and y
- Solve for y
- Write the result as f⁻¹(x)

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(a)
Given:
f(x) = (x + 3)/4

Step 1: Let y = (x + 3)/4
Step 2: Swap x and y → x = (y + 3)/4
Step 3: Solve for y
Multiply both sides by 4:
4x = y + 3
Subtract 3:
y = 4x - 3

So, inverse is:
f⁻¹(x) = 4x - 3

✔ Check: Plug in x=1 → f(1)=(1+3)/4=1 → f¹(1)=4(1)-3=1 ✔️
Plug in x=5 → f(5)=(5+3)/4=2 → f⁻¹(2)=8-3=5 ✔️

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(b)
Given:
g(x) = (x - 2) [Note: domain x ≥ 2]

Step 1: Let y = √(x - 2)
Step 2: Swap x and y → x = (y - 2)
Step 3: Solve for y
Square both sides:
x² = y - 2
Add 2:
y = x² + 2

But note: since original function outputs only non-negative values (square root), the inverse must have domain x ≥ 0.

So, inverse is:
g¹(x) = x² + 2, for x ≥ 0

✔ Check: g(6) = √(6-2) = √4 = 2 → g⁻¹(2) = 4 + 2 = 6 ✔️
g(3) = √(3-2) = 1 → g⁻¹(1) = 1 + 2 = 3 ✔️

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(c)
Given:
h(x) = 2/(x + 1) [x ≠ -1]

Step 1: Let y = 2/(x + 1)
Step 2: Swap x and y → x = 2/(y + 1)
Step 3: Solve for y
Multiply both sides by (y + 1):
x(y + 1) = 2
Divide both sides by x (x ≠ 0):
y + 1 = 2/x
Subtract 1:
y = (2/x) - 1

So, inverse is:
h⁻¹(x) = (2/x) - 1, for x ≠ 0

✔ Check: h(1) = 2/(1+1) = 1 → h⁻¹(1) = 2/1 - 1 = 1 ✔️
h(3) = 2/(3+1) = 0.5 → h⁻¹(0.5) = 2/0.5 - 1 = 4 - 1 = 3 ✔️

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Problem 2: Find the inverse of each relation.

A relation is just a set of ordered pairs. To find the inverse, swap x and y in each pair.

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(a)
Relation: {(1, 2), (3, 4), (5, 6)}

Swap each pair:
→ {(2, 1), (4, 3), (6, 5)}

Answer: {(2, 1), (4, 3), (6, 5)}

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(b)
Relation: {(-2, 7), (0, 3), (4, -1)}

Swap each pair:
→ {(7, -2), (3, 0), (-1, 4)}

Answer: {(7, -2), (3, 0), (-1, 4)}

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(c)
Relation: {(½, ¼), (¾, ⅛), (, ⅓)}

Swap each pair:
→ {(¼, ½), (, ¾), (⅓, ⅔)}

Answer: {(¼, ½), (⅛, ¾), (⅓, ⅔)}

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(d)
Relation: {(√2, π), (e, ln 2), (log₃9, 2)}

First simplify where possible:

- log₃9 = log₃(3²) = 2 → so last pair is (2, 2)

Now swap all:

→ {(π, 2), (ln 2, e), (2, 2)}

Answer: {(π, √2), (ln 2, e), (2, 2)}

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Final Answer:

1(a) f⁻¹(x) = 4x - 3
1(b) g⁻¹(x) = x² + 2, x ≥ 0
1(c) h⁻¹(x) = (2/x) - 1, x ≠ 0

2(a) {(2, 1), (4, 3), (6, 5)}
2(b) {(7, -2), (3, 0), (-1, 4)}
2(c) {(¼, ½), (, ¾), (⅓, ⅔)}
2(d) {(π, √2), (ln 2, e), (2, 2)}