Free Printable Inverse Functions Worksheets for Students - Free Printable

Worksheet 7.4: Inverse Functions

#### Inverse Relations
Find the inverse for each relation.

1. Relation: \(\{(1, -3), (-2, 3), (5, 1), (6, 4)\}\)

To find the inverse of a relation, swap the \(x\)-coordinates and \(y\)-coordinates of each ordered pair.
\[
\text{Inverse: } \{(-3, 1), (3, -2), (1, 5), (4, 6)\}
\]

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

Similarly, swap the coordinates:
\[
\text{Inverse: } \{(7, -5), (-8, -6), (-2, 1), (3, 10)\}
\]

---

#### Finding Inverses
Find an equation for the inverse for each of the following relations.

3. Relation: \(y = 3x + 2\)

To find the inverse, solve for \(x\) in terms of \(y\):
\[
y = 3x + 2
\]
Subtract 2 from both sides:
\[
y - 2 = 3x
\]
Divide by 3:
\[
x = \frac{y - 2}{3}
\]
Swap \(x\) and \(y\) to get the inverse function:
\[
y = \frac{x - 2}{3}
\]
\[
\boxed{y = \frac{x - 2}{3}}
\]

4. Relation: \(y = -5x - 7\)

Solve for \(x\) in terms of \(y\):
\[
y = -5x - 7
\]
Add 7 to both sides:
\[
y + 7 = -5x
\]
Divide by \(-5\):
\[
x = -\frac{y + 7}{5}
\]
Swap \(x\) and \(y\):
\[
y = -\frac{x + 7}{5}
\]
\[
\boxed{y = -\frac{x + 7}{5}}
\]

5. Relation: \(y = 12x - 3\)

Solve for \(x\) in terms of \(y\):
\[
y = 12x - 3
\]
Add 3 to both sides:
\[
y + 3 = 12x
\]
Divide by 12:
\[
x = \frac{y + 3}{12}
\]
Swap \(x\) and \(y\):
\[
y = \frac{x + 3}{12}
\]
\[
\boxed{y = \frac{x + 3}{12}}
\]

6. Relation: \(y = -8x + 16\)

Solve for \(x\) in terms of \(y\):
\[
y = -8x + 16
\]
Subtract 16 from both sides:
\[
y - 16 = -8x
\]
Divide by \(-8\):
\[
x = -\frac{y - 16}{8}
\]
Swap \(x\) and \(y\):
\[
y = -\frac{x - 16}{8}
\]
\[
\boxed{y = -\frac{x - 16}{8}}
\]

7. Relation: \(y = \frac{2}{3}x - 5\)

Solve for \(x\) in terms of \(y\):
\[
y = \frac{2}{3}x - 5
\]
Add 5 to both sides:
\[
y + 5 = \frac{2}{3}x
\]
Multiply by \(\frac{3}{2}\):
\[
x = \frac{3}{2}(y + 5)
\]
Swap \(x\) and \(y\):
\[
y = \frac{3}{2}(x + 5)
\]
\[
\boxed{y = \frac{3}{2}(x + 5)}
\]

8. Relation: \(y = -\frac{3}{4}x + 5\)

Solve for \(x\) in terms of \(y\):
\[
y = -\frac{3}{4}x + 5
\]
Subtract 5 from both sides:
\[
y - 5 = -\frac{3}{4}x
\]
Multiply by \(-\frac{4}{3}\):
\[
x = -\frac{4}{3}(y - 5)
\]
Swap \(x\) and \(y\):
\[
y = -\frac{4}{3}(x - 5)
\]
\[
\boxed{y = -\frac{4}{3}(x - 5)}
\]

9. Relation: \(y = -\frac{5}{8}x + 10\)

Solve for \(x\) in terms of \(y\):
\[
y = -\frac{5}{8}x + 10
\]
Subtract 10 from both sides:
\[
y - 10 = -\frac{5}{8}x
\]
Multiply by \(-\frac{8}{5}\):
\[
x = -\frac{8}{5}(y - 10)
\]
Swap \(x\) and \(y\):
\[
y = -\frac{8}{5}(x - 10)
\]
\[
\boxed{y = -\frac{8}{5}(x - 10)}
\]

10. Relation: \(y = \frac{1}{2}x + 8\)

Solve for \(x\) in terms of \(y\):
\[
y = \frac{1}{2}x + 8
\]
Subtract 8 from both sides:
\[
y - 8 = \frac{1}{2}x
\]
Multiply by 2:
\[
x = 2(y - 8)
\]
Swap \(x\) and \(y\):
\[
y = 2(x - 8)
\]
\[
\boxed{y = 2(x - 8)}
\]

11. Relation: \(y = x^2 + 5\)

Solve for \(x\) in terms of \(y\):
\[
y = x^2 + 5
\]
Subtract 5 from both sides:
\[
y - 5 = x^2
\]
Take the square root of both sides:
\[
x = \pm\sqrt{y - 5}
\]
Since the original function \(y = x^2 + 5\) is not one-to-one, the inverse is not a function unless we restrict the domain. Assuming the principal (positive) square root:
\[
x = \sqrt{y - 5}
\]
Swap \(x\) and \(y\):
\[
y = \sqrt{x - 5}
\]
\[
\boxed{y = \sqrt{x - 5}}
\]

12. Relation: \(y = x^2 - 4\)

Solve for \(x\) in terms of \(y\):
\[
y = x^2 - 4
\]
Add 4 to both sides:
\[
y + 4 = x^2
\]
Take the square root of both sides:
\[
x = \pm\sqrt{y + 4}
\]
Assuming the principal (positive) square root:
\[
x = \sqrt{y + 4}
\]
Swap \(x\) and \(y\):
\[
y = \sqrt{x + 4}
\]
\[
\boxed{y = \sqrt{x + 4}}
\]

13. Relation: \(y = (x + 3)^2\)

Solve for \(x\) in terms of \(y\):
\[
y = (x + 3)^2
\]
Take the square root of both sides:
\[
\sqrt{y} = x + 3
\]
Subtract 3 from both sides:
\[
x = \sqrt{y} - 3
\]
Swap \(x\) and \(y\):
\[
y = \sqrt{x} - 3
\]
\[
\boxed{y = \sqrt{x} - 3}
\]

14. Relation: \(y = (x - 6)^2\)

Solve for \(x\) in terms of \(y\):
\[
y = (x - 6)^2
\]
Take the square root of both sides:
\[
\sqrt{y} = x - 6
\]
Add 6 to both sides:
\[
x = \sqrt{y} + 6
\]
Swap \(x\) and \(y\):
\[
y = \sqrt{x} + 6
\]
\[
\boxed{y = \sqrt{x} + 6}
\]

15. Relation: \(y = \sqrt{x - 2}, \, y \geq 0\)

Solve for \(x\) in terms of \(y\):
\[
y = \sqrt{x - 2}
\]
Square both sides:
\[
y^2 = x - 2
\]
Add 2 to both sides:
\[
x = y^2 + 2
\]
Swap \(x\) and \(y\):
\[
y = x^2 + 2
\]
\[
\boxed{y = x^2 + 2}
\]

16. Relation: \(y = \sqrt{x + 5}, \, y \geq 0\)

Solve for \(x\) in terms of \(y\):
\[
y = \sqrt{x + 5}
\]
Square both sides:
\[
y^2 = x + 5
\]
Subtract 5 from both sides:
\[
x = y^2 - 5
\]
Swap \(x\) and \(y\):
\[
y = x^2 - 5
\]
\[
\boxed{y = x^2 - 5}
\]

17. Relation: \(y = \sqrt{x} + 8, \, y \geq 8\)

Solve for \(x\) in terms of \(y\):
\[
y = \sqrt{x} + 8
\]
Subtract 8 from both sides:
\[
y - 8 = \sqrt{x}
\]
Square both sides:
\[
(y - 8)^2 = x
\]
Swap \(x\) and \(y\):
\[
y = (x - 8)^2
\]
\[
\boxed{y = (x - 8)^2}
\]

18. Relation: \(y = \sqrt{x} - 7, \, y \geq -7\)

Solve for \(x\) in terms of \(y\):
\[
y = \sqrt{x} - 7
\]
Add 7 to both sides:
\[
y + 7 = \sqrt{x}
\]
Square both sides:
\[
(y + 7)^2 = x
\]
Swap \(x\) and \(y\):
\[
y = (x + 7)^2
\]
\[
\boxed{y = (x + 7)^2}
\]

---

#### Verifying Inverses
Verify that \(f\) and \(g\) are inverse functions.

19. Functions: \(f(x) = x + 6\), \(g(x) = x - 6\)

To verify, check if \(f(g(x)) = x\) and \(g(f(x)) = x\):
\[
f(g(x)) = f(x - 6) = (x - 6) + 6 = x
\]
\[
g(f(x)) = g(x + 6) = (x + 6) - 6 = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

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

Check \(f(g(x))\) and \(g(f(x))\):
\[
f(g(x)) = f\left(\frac{x - 2}{5}\right) = 5\left(\frac{x - 2}{5}\right) + 2 = x - 2 + 2 = x
\]
\[
g(f(x)) = g(5x + 2) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

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

Check \(f(g(x))\) and \(g(f(x))\):
\[
f(g(x)) = f\left(-\frac{1}{3}x - 3\right) = -3\left(-\frac{1}{3}x - 3\right) - 9 = x + 9 - 9 = x
\]
\[
g(f(x)) = g(-3x - 9) = -\frac{1}{3}(-3x - 9) - 3 = x + 3 - 3 = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

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

Check \(f(g(x))\) and \(g(f(x))\):
\[
f(g(x)) = f\left(\frac{x + 7}{2}\right) = 2\left(\frac{x + 7}{2}\right) - 7 = x + 7 - 7 = x
\]
\[
g(f(x)) = g(2x - 7) = \frac{(2x - 7) + 7}{2} = \frac{2x}{2} = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

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

Check \(f(g(x))\) and \(g(f(x))\):
\[
f(g(x)) = f\left(-\frac{1}{4}x + 2\right) = -4\left(-\frac{1}{4}x + 2\right) + 8 = x - 8 + 8 = x
\]
\[
g(f(x)) = g(-4x + 8) = -\frac{1}{4}(-4x + 8) + 2 = x - 2 + 2 = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

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

Check \(f(g(x))\) and \(g(f(x))\):
\[
f(g(x)) = f(2x + 14) = \frac{1}{2}(2x + 14) - 7 = x + 7 - 7 = x
\]
\[
g(f(x)) = g\left(\frac{1}{2}x - 7\right) = 2\left(\frac{1}{2}x - 7\right) + 14 = x - 14 + 14 = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.
\[
\boxed{\text{Yes}}
\]

---

Final Answers:

- 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. \(y = -\frac{x - 16}{8}\)
7. \(y = \frac{3}{2}(x + 5)\)
8. \(y = -\frac{4}{3}(x - 5)\)
9. \(y = -\frac{8}{5}(x - 10)\)
10. \(y = 2(x - 8)\)
11. \(y = \sqrt{x - 5}\)
12. \(y = \sqrt{x + 4}\)
13. \(y = \sqrt{x} - 3\)
14. \(y = \sqrt{x} + 6\)
15. \(y = x^2 + 2\)
16. \(y = x^2 - 5\)
17. \(y = (x - 8)^2\)
18. \(y = (x + 7)^2\)

- Verifying Inverses:
19. \(\boxed{\text{Yes}}\)
20. \(\boxed{\text{Yes}}\)
21. \(\boxed{\text{Yes}}\)
22. \(\boxed{\text{Yes}}\)
23. \(\boxed{\text{Yes}}\)
24. \(\boxed{\text{Yes}}\)