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 \(x\)- and \(y\)-coordinates of each ordered pair.
- Original: \((1, -3)\) → Inverse: \((-3, 1)\)
- Original: \((-2, 3)\) → Inverse: \((3, -2)\)
- Original: \((5, 1)\) → Inverse: \((1, 5)\)
- Original: \((6, 4)\) → Inverse: \((4, 6)\)

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

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

Similarly, swap the \(x\)- and \(y\)-coordinates:
- Original: \((-5, 7)\) → Inverse: \((7, -5)\)
- Original: \((-6, -8)\) → Inverse: \((-8, -6)\)
- Original: \((1, -2)\) → Inverse: \((-2, 1)\)
- Original: \((10, 3)\) → Inverse: \((3, 10)\)

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

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#### 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:
\[
y = \frac{x - 2}{3}
\]

Inverse Equation: \(y = \frac{x - 2}{3}\)

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 \(y\):
\[
y = -\frac{x + 7}{5}
\]

Inverse Equation: \(y = -\frac{x + 7}{5}\)

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 \(y\):
\[
y = \frac{x + 3}{12}
\]

Inverse Equation: \(y = \frac{x + 3}{12}\)

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 \(y\):
\[
y = -\frac{x - 16}{8}
\]

Inverse Equation: \(y = -\frac{x - 16}{8}\)

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 \(y\):
\[
y = \frac{3}{2}(x + 5)
\]

Inverse Equation: \(y = \frac{3}{2}(x + 5)\)

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 \(y\):
\[
y = -\frac{4}{3}(x - 5)
\]

Inverse Equation: \(y = -\frac{4}{3}(x - 5)\)

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 \(y\):
\[
y = -\frac{8}{5}(x - 10)
\]

Inverse Equation: \(y = -\frac{8}{5}(x - 10)\)

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 \(y\):
\[
y = 2(x - 8)
\]

Inverse Equation: \(y = 2(x - 8)\)

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 (it fails the horizontal line test), we need to restrict the domain to make it invertible. Assuming the principal (positive) square root:
\[
x = \sqrt{y - 5}
\]
Swap \(x\) and \(y\):
\[
y = \sqrt{x - 5}
\]

Inverse Equation: \(y = \sqrt{x - 5}\) (with domain \(x \geq 5\))

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 \(y\):
\[
y = \sqrt{x + 4}
\]

Inverse Equation: \(y = \sqrt{x + 4}\) (with domain \(x \geq -4\))

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 \(y\):
\[
y = \sqrt{x} - 3
\]

Inverse Equation: \(y = \sqrt{x} - 3\) (with domain \(x \geq 0\))

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 \(y\):
\[
y = \sqrt{x} + 6
\]

Inverse Equation: \(y = \sqrt{x} + 6\) (with domain \(x \geq 0\))

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 \(y\):
\[
y = x^2 + 2
\]

Inverse Equation: \(y = x^2 + 2\) (with domain \(x \geq 0\))

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 \(y\):
\[
y = x^2 - 5
\]

Inverse Equation: \(y = x^2 - 5\) (with domain \(x \geq 0\))

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 \(y\):
\[
y = (x - 8)^2
\]

Inverse Equation: \(y = (x - 8)^2\) (with domain \(x \geq 8\))

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 \(y\):
\[
y = (x + 7)^2
\]

Inverse Equation: \(y = (x + 7)^2\) (with domain \(x \geq -7\))

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

To verify that \(f\) and \(g\) are inverses, check if:
\[
(f \circ g)(x) = x \quad \text{and} \quad (g \circ f)(x) = x
\]

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

Compute \((f \circ g)(x)\):
\[
(f \circ g)(x) = f(g(x)) = f(x - 6) = (x - 6) + 6 = x
\]
Compute \((g \circ f)(x)\):
\[
(g \circ f)(x) = g(f(x)) = g(x + 6) = (x + 6) - 6 = x
\]
Since both conditions are satisfied, \(f\) and \(g\) are inverses.

Verification: Yes

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

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

Verification: Yes

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

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

Verification: Yes

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

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

Verification: Yes

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

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

Verification: Yes

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

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

Verification: Yes

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Final Answers:

1. \(\boxed{\{(-3, 1), (3, -2), (1, 5), (4, 6)\}}\)
2. \(\boxed{\{(7, -5), (-8, -6), (-2, 1), (3, 10)\}}\)
3. \(\boxed{y = \frac{x - 2}{3}}\)
4. \(\boxed{y = -\frac{x + 7}{5}}\)
5. \(\boxed{y = \frac{x + 3}{12}}\)
6. \(\boxed{y = -\frac{x - 16}{8}}\)
7. \(\boxed{y = \frac{3}{2}(x + 5)}\)
8. \(\boxed{y = -\frac{4}{3}(x - 5)}\)
9. \(\boxed{y = -\frac{8}{5}(x - 10)}\)
10. \(\boxed{y = 2(x - 8)}\)
11. \(\boxed{y = \sqrt{x - 5}}\)
12. \(\boxed{y = \sqrt{x + 4}}\)
13. \(\boxed{y = \sqrt{x} - 3}\)
14. \(\boxed{y = \sqrt{x} + 6}\)
15. \(\boxed{y = x^2 + 2}\)
16. \(\boxed{y = x^2 - 5}\)
17. \(\boxed{y = (x - 8)^2}\)
18. \(\boxed{y = (x + 7)^2}\)
19. \(\boxed{\text{Yes}}\)
20. \(\boxed{\text{Yes}}\)
21. \(\boxed{\text{Yes}}\)
22. \(\boxed{\text{Yes}}\)
23. \(\boxed{\text{Yes}}\)
24. \(\boxed{\text{Yes}}\)