worksheet 7.4 inverse functions | Lecture notes Elementary ... - Free Printable

Here are the solutions for the worksheet problems, broken down by section.

Inverse Relations

To find the inverse of a relation given as a set of points, you simply swap the $x$ and $y$ coordinates for every point. $(x, y)$ becomes $(y, x)$.

1. Original: $\{ (1, -3), (-2, 3), (5, 1), (6, 4) \}$
* Swap $(1, -3) \rightarrow (-3, 1)$
* Swap $(-2, 3) \rightarrow (3, -2)$
* Swap $(5, 1) \rightarrow (1, 5)$
* Swap $(6, 4) \rightarrow (4, 6)$
Answer: $\{ (-3, 1), (3, -2), (1, 5), (4, 6) \}$

2. Original: $\{ (-5, 7), (-6, -8), (1, -2), (10, 3) \}$
* Swap $(-5, 7) \rightarrow (7, -5)$
* Swap $(-6, -8) \rightarrow (-8, -6)$
* Swap $(1, -2) \rightarrow (-2, 1)$
* Swap $(10, 3) \rightarrow (3, 10)$
Answer: $\{ (7, -5), (-8, -6), (-2, 1), (3, 10) \}$

---

Finding Inverses

To find the equation of an inverse function:
1. Replace $y$ with $x$ and $x$ with $y$.
2. Solve the new equation for $y$.

3. $y = 3x + 2$
* Switch: $x = 3y + 2$
* Subtract 2: $x - 2 = 3y$
* Divide by 3: $y = \frac{x - 2}{3}$

4. $y = -5x - 7$
* Switch: $x = -5y - 7$
* Add 7: $x + 7 = -5y$
* Divide by -5: $y = \frac{x + 7}{-5}$ or $y = -\frac{x + 7}{5}$

5. $y = 12x - 3$
* Switch: $x = 12y - 3$
* Add 3: $x + 3 = 12y$
* Divide by 12: $y = \frac{x + 3}{12}$

6. $y = -8x + 16$
* Switch: $x = -8y + 16$
* Subtract 16: $x - 16 = -8y$
* Divide by -8: $y = \frac{x - 16}{-8}$ which simplifies to $y = -\frac{1}{8}x + 2$

7. $y = \frac{2}{3}x - 5$
* Switch: $x = \frac{2}{3}y - 5$
* Add 5: $x + 5 = \frac{2}{3}y$
* Multiply by $\frac{3}{2}$ (reciprocal): $y = \frac{3}{2}(x + 5)$ or $y = \frac{3x + 15}{2}$

8. $y = -\frac{3}{4}x + 5$
* Switch: $x = -\frac{3}{4}y + 5$
* Subtract 5: $x - 5 = -\frac{3}{4}y$
* Multiply by $-\frac{4}{3}$: $y = -\frac{4}{3}(x - 5)$ or $y = \frac{-4x + 20}{3}$

9. $y = -\frac{5}{8}x + 10$
* Switch: $x = -\frac{5}{8}y + 10$
* Subtract 10: $x - 10 = -\frac{5}{8}y$
* Multiply by $-\frac{8}{5}$: $y = -\frac{8}{5}(x - 10)$ or $y = \frac{-8x + 80}{5}$

10. $y = \frac{1}{2}x + 8$
* Switch: $x = \frac{1}{2}y + 8$
* Subtract 8: $x - 8 = \frac{1}{2}y$
* Multiply by 2: $y = 2(x - 8)$ or $y = 2x - 16$

11. $y = x^2 + 5$
* Switch: $x = y^2 + 5$
* Subtract 5: $x - 5 = y^2$
* Square root: $y = \pm\sqrt{x - 5}$

12. $y = x^2 - 4$
* Switch: $x = y^2 - 4$
* Add 4: $x + 4 = y^2$
* Square root: $y = \pm\sqrt{x + 4}$

13. $y = (x + 3)^2$
* Switch: $x = (y + 3)^2$
* Square root: $\pm\sqrt{x} = y + 3$
* Subtract 3: $y = -3 \pm \sqrt{x}$

14. $y = (x - 6)^2$
* Switch: $x = (y - 6)^2$
* Square root: $\pm\sqrt{x} = y - 6$
* Add 6: $y = 6 \pm \sqrt{x}$

15. $y = \sqrt{x - 2}, y \ge 0$
* Switch: $x = \sqrt{y - 2}$
* Square both sides: $x^2 = y - 2$
* Add 2: $y = x^2 + 2$
* *Note on domain:* Since original $y \ge 0$, the new $x \ge 0$.

16. $y = \sqrt{x + 5}, y \ge 0$
* Switch: $x = \sqrt{y + 5}$
* Square both sides: $x^2 = y + 5$
* Subtract 5: $y = x^2 - 5$
* *Note on domain:* New $x \ge 0$.

17. $y = \sqrt{x} + 8, y \ge 8$
* Switch: $x = \sqrt{y} + 8$
* Subtract 8: $x - 8 = \sqrt{y}$
* Square both sides: $y = (x - 8)^2$
* *Note on domain:* Since original $y \ge 8$, the new $x \ge 8$.

18. $y = \sqrt{x} - 7, y \ge -7$
* Switch: $x = \sqrt{y} - 7$
* Add 7: $x + 7 = \sqrt{y}$
* Square both sides: $y = (x + 7)^2$
* *Note on domain:* Since original $y \ge -7$, the new $x \ge -7$.

---

Verifying Inverses

To verify that $f(x)$ and $g(x)$ are inverses, you must show that $f(g(x)) = x$ AND $g(f(x)) = x$.

19. $f(x) = x + 6$, $g(x) = x - 6$
* $f(g(x)) = (x - 6) + 6 = x$
* $g(f(x)) = (x + 6) - 6 = x$
* Yes, they are inverses.

20. $f(x) = 5x + 2$, $g(x) = \frac{x - 2}{5}$
* $f(g(x)) = 5(\frac{x - 2}{5}) + 2 = (x - 2) + 2 = x$
* $g(f(x)) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x$
* Yes, they are inverses.

21. $f(x) = -3x - 9$, $g(x) = -\frac{1}{3}x - 3$
* $f(g(x)) = -3(-\frac{1}{3}x - 3) - 9 = (x + 9) - 9 = x$
* $g(f(x)) = -\frac{1}{3}(-3x - 9) - 3 = (x + 3) - 3 = x$
* Yes, they are inverses.

22. $f(x) = 2x - 7$, $g(x) = \frac{x + 7}{2}$
* $f(g(x)) = 2(\frac{x + 7}{2}) - 7 = (x + 7) - 7 = x$
* $g(f(x)) = \frac{(2x - 7) + 7}{2} = \frac{2x}{2} = x$
* Yes, they are inverses.

23. $f(x) = -4x + 8$, $g(x) = -\frac{1}{4}x + 2$
* $f(g(x)) = -4(-\frac{1}{4}x + 2) + 8 = (x - 8) + 8 = x$
* $g(f(x)) = -\frac{1}{4}(-4x + 8) + 2 = (x - 2) + 2 = x$
* Yes, they are inverses.

24. $f(x) = \frac{1}{2}x - 7$, $g(x) = 2x + 14$
* $f(g(x)) = \frac{1}{2}(2x + 14) - 7 = (x + 7) - 7 = x$
* $g(f(x)) = 2(\frac{1}{2}x - 7) + 14 = (x - 14) + 14 = x$
* Yes, they are inverses.

Final Answer:
1. $\{ (-3, 1), (3, -2), (1, 5), (4, 6) \}$
2. $\{ (7, -5), (-8, -6), (-2, 1), (3, 10) \}$
3. $y = \frac{x - 2}{3}$
4. $y = -\frac{x + 7}{5}$
5. $y = \frac{x + 3}{12}$
6. $y = -\frac{1}{8}x + 2$
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. $y = x^2 + 2, x \ge 0$
16. $y = x^2 - 5, x \ge 0$
17. $y = (x - 8)^2, x \ge 8$
18. $y = (x + 7)^2, x \ge -7$
19-24. All pairs are verified as inverse functions.