Free Printable Inverse Functions Worksheets for Students - Free Printable

Let's solve each part of Worksheet 7.4: Inverse Functions step by step, with explanations.

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Part 1: Inverse Relations

To find the inverse of a relation, swap the x and y values in each ordered pair.

#### 1. $\{(1, -3), (-2, 3), (5, 1), (6, 4)\}$

Swap each pair:
- $(1, -3) \rightarrow (-3, 1)$
- $(-2, 3) \rightarrow (3, -2)$
- $(5, 1) \rightarrow (1, 5)$
- $(6, 4) \rightarrow (4, 6)$

✔ Inverse: $\{(-3, 1), (3, -2), (1, 5), (4, 6)\}$

#### 2. $\{(-5, 7), (-6, -8), (1, -2), (10, 3)\}$

Swap each pair:
- $(-5, 7) \rightarrow (7, -5)$
- $(-6, -8) \rightarrow (-8, -6)$
- $(1, -2) \rightarrow (-2, 1)$
- $(10, 3) \rightarrow (3, 10)$

✔ Inverse: $\{(7, -5), (-8, -6), (-2, 1), (3, 10)\}$

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Part 2: Finding Inverses

To find the inverse of a function $y = f(x)$, follow these steps:

1. Replace $y$ with $x$, and $x$ with $y$: $x = f(y)$
2. Solve for $y$
3. The result is the inverse function $f^{-1}(x)$

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#### 3. $y = 3x + 2$

Swap $x$ and $y$:
$x = 3y + 2$

Solve for $y$:
$x - 2 = 3y$
$y = \frac{x - 2}{3}$

✔ Inverse: $y = \frac{x - 2}{3}$

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#### 4. $y = -5x - 7$

Swap:
$x = -5y - 7$

Solve:
$x + 7 = -5y$
$y = -\frac{x + 7}{5}$

✔ Inverse: $y = -\frac{x + 7}{5}$

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#### 5. $y = 12x - 3$

Swap:
$x = 12y - 3$

Solve:
$x + 3 = 12y$
$y = \frac{x + 3}{12}$

✔ Inverse: $y = \frac{x + 3}{12}$

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#### 6. $y = -8x + 16$

Swap:
$x = -8y + 16$

Solve:
$x - 16 = -8y$
$y = \frac{16 - x}{8}$

✔ Inverse: $y = \frac{16 - x}{8}$ or $y = -\frac{1}{8}x + 2$

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#### 7. $y = \frac{2}{3}x - 5$

Swap:
$x = \frac{2}{3}y - 5$

Solve:
$x + 5 = \frac{2}{3}y$
Multiply both sides by 3: $3x + 15 = 2y$
$y = \frac{3x + 15}{2}$

✔ Inverse: $y = \frac{3x + 15}{2}$

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#### 8. $y = -\frac{3}{4}x + 5$

Swap:
$x = -\frac{3}{4}y + 5$

Solve:
$x - 5 = -\frac{3}{4}y$
Multiply both sides by 4: $4x - 20 = -3y$
$y = \frac{-4x + 20}{3}$

✔ Inverse: $y = \frac{-4x + 20}{3}$

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#### 9. $y = -\frac{5}{8}x + 10$

Swap:
$x = -\frac{5}{8}y + 10$

Solve:
$x - 10 = -\frac{5}{8}y$
Multiply by 8: $8x - 80 = -5y$
$y = \frac{-8x + 80}{5}$

✔ Inverse: $y = \frac{-8x + 80}{5}$

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#### 10. $y = \frac{1}{2}x + 8$

Swap:
$x = \frac{1}{2}y + 8$

Solve:
$x - 8 = \frac{1}{2}y$
$y = 2(x - 8) = 2x - 16$

✔ Inverse: $y = 2x - 16$

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#### 11. $y = x^2 + 5$

This is a quadratic, so it's not one-to-one unless domain is restricted.

But since no restriction is given, we cannot define an inverse function over all real numbers because it fails the horizontal line test.

However, if we consider the relation, we can still find the inverse relation.

Swap:
$x = y^2 + 5$

Solve:
$y^2 = x - 5$
$y = \pm \sqrt{x - 5}$

So the inverse relation is $y = \pm \sqrt{x - 5}$, but it’s not a function unless we restrict domain.

✔ Inverse relation: $y = \pm \sqrt{x - 5}$

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#### 12. $y = x^2 - 4$

Same logic as above.

Swap:
$x = y^2 - 4$
$y^2 = x + 4$
$y = \pm \sqrt{x + 4}$

✔ Inverse relation: $y = \pm \sqrt{x + 4}$

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#### 13. $y = (x + 3)^2$

Swap:
$x = (y + 3)^2$

Solve:
$\sqrt{x} = |y + 3|$ → two cases

But again, unless domain is restricted, inverse is not a function.

We get: $y + 3 = \pm \sqrt{x}$
$y = -3 \pm \sqrt{x}$

✔ Inverse relation: $y = -3 \pm \sqrt{x}$

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#### 14. $y = (x - 6)^2$

Swap:
$x = (y - 6)^2$
$y - 6 = \pm \sqrt{x}$
$y = 6 \pm \sqrt{x}$

✔ Inverse relation: $y = 6 \pm \sqrt{x}$

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#### 15. $y = \sqrt{x - 2},\ y \geq 0$

Domain: $x \geq 2$, Range: $y \geq 0$

Swap:
$x = \sqrt{y - 2}$

Now solve for $y$:
Square both sides: $x^2 = y - 2$
$y = x^2 + 2$

But since original $y \geq 0$, and $x = \sqrt{y - 2} \geq 0$, so $x \geq 0$

So inverse is defined for $x \geq 0$

✔ Inverse: $y = x^2 + 2,\ x \geq 0$

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#### 16. $y = \sqrt{x + 5},\ y \geq 0$

Domain: $x \geq -5$, Range: $y \geq 0$

Swap:
$x = \sqrt{y + 5}$

Square: $x^2 = y + 5$
$y = x^2 - 5$

And since $x = \sqrt{y + 5} \geq 0$, then $x \geq 0$

✔ Inverse: $y = x^2 - 5,\ x \geq 0$

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#### 17. $y = \sqrt{x} + 8,\ y \geq 8$

Domain: $x \geq 0$, Range: $y \geq 8$

Swap:
$x = \sqrt{y} + 8$

Solve:
$x - 8 = \sqrt{y}$
$(x - 8)^2 = y$

Since $y \geq 8$, then $x = \sqrt{y} + 8 \geq 8$, so $x \geq 8$

✔ Inverse: $y = (x - 8)^2,\ x \geq 8$

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#### 18. $y = \sqrt{x} - 7,\ y \geq -7$

Domain: $x \geq 0$, Range: $y \geq -7$

Swap:
$x = \sqrt{y} - 7$

Solve:
$x + 7 = \sqrt{y}$
$(x + 7)^2 = y$

Since $y \geq -7$, then $x = \sqrt{y} - 7 \geq -7$, so $x \geq -7$

But also, $\sqrt{y} \geq 0$, so $x + 7 \geq 0 \Rightarrow x \geq -7$

✔ Inverse: $y = (x + 7)^2,\ x \geq -7$

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Part 3: Verifying Inverses

Two functions $f$ and $g$ are inverses if:
$$
f(g(x)) = x \quad \text{and} \quad g(f(x)) = x
$$

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#### 19. $f(x) = x + 6$, $g(x) = x - 6$

Check:
- $f(g(x)) = f(x - 6) = (x - 6) + 6 = x$
- $g(f(x)) = g(x + 6) = (x + 6) - 6 = x$

✔ Yes, they are inverses

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#### 20. $f(x) = 5x + 2$, $g(x) = \frac{x - 2}{5}$

- $f(g(x)) = f\left(\frac{x - 2}{5}\right) = 5\left(\frac{x - 2}{5}\right) + 2 = (x - 2) + 2 = x$
- $g(f(x)) = g(5x + 2) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x$

✔ Yes, inverses

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#### 21. $f(x) = -3x - 9$, $g(x) = -\frac{1}{3}x - 3$

- $f(g(x)) = f\left(-\frac{1}{3}x - 3\right) = -3\left(-\frac{1}{3}x - 3\right) - 9 = (x + 9) - 9 = x$
- $g(f(x)) = g(-3x - 9) = -\frac{1}{3}(-3x - 9) - 3 = (x + 3) - 3 = x$

✔ Yes, inverses

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#### 22. $f(x) = 2x - 7$, $g(x) = \frac{x + 7}{2}$

- $f(g(x)) = f\left(\frac{x + 7}{2}\right) = 2\left(\frac{x + 7}{2}\right) - 7 = (x + 7) - 7 = x$
- $g(f(x)) = g(2x - 7) = \frac{(2x - 7) + 7}{2} = \frac{2x}{2} = x$

✔ Yes, inverses

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#### 23. $f(x) = -4x + 8$, $g(x) = -\frac{1}{4}x + 2$

- $f(g(x)) = f\left(-\frac{1}{4}x + 2\right) = -4\left(-\frac{1}{4}x + 2\right) + 8 = (x - 8) + 8 = x$
- $g(f(x)) = g(-4x + 8) = -\frac{1}{4}(-4x + 8) + 2 = (x - 2) + 2 = x$

✔ Yes, inverses

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#### 24. $f(x) = \frac{1}{2}x - 7$, $g(x) = 2x + 14$

- $f(g(x)) = f(2x + 14) = \frac{1}{2}(2x + 14) - 7 = (x + 7) - 7 = x$
- $g(f(x)) = g\left(\frac{1}{2}x - 7\right) = 2\left(\frac{1}{2}x - 7\right) + 14 = (x - 14) + 14 = x$

✔ Yes, inverses

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✔ Final Answers Summary:

#### Inverse Relations
1. $\{(-3, 1), (3, -2), (1, 5), (4, 6)\}$
2. $\{(7, -5), (-8, -6), (-2, 1), (3, 10)\}$

#### Finding Inverses
3. $y = \frac{x - 2}{3}$
4. $y = -\frac{x + 7}{5}$
5. $y = \frac{x + 3}{12}$
6. $y = \frac{16 - x}{8}$
7. $y = \frac{3x + 15}{2}$
8. $y = \frac{-4x + 20}{3}$
9. $y = \frac{-8x + 80}{5}$
10. $y = 2x - 16$
11. $y = \pm \sqrt{x - 5}$
12. $y = \pm \sqrt{x + 4}$
13. $y = -3 \pm \sqrt{x}$
14. $y = 6 \pm \sqrt{x}$
15. $y = x^2 + 2,\ x \geq 0$
16. $y = x^2 - 5,\ x \geq 0$
17. $y = (x - 8)^2,\ x \geq 8$
18. $y = (x + 7)^2,\ x \geq -7$

#### Verifying Inverses
19–24: All are inverses ✔

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