Worksheet 7.4 Inverse Functions Inverse Relations: Continued On ... - Free Printable

Let's solve the problems on Worksheet 7.4: Inverse Functions step by step, with explanations.

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Part 1: Inverse Relations

We are given sets of ordered pairs and asked to find their inverses.

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

To find the inverse, swap the $x$ and $y$ values in each pair.

- $(1, -3)$ → $(-3, 1)$
- $(-2, 3)$ → $(3, -2)$
- $(5, 1)$ → $(1, 5)$
- $(6, 4)$ → $(4, 6)$

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

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

Swap $x$ and $y$:

- $(-5, 7)$ → $(7, -5)$
- $(-6, -8)$ → $(-8, -6)$
- $(1, -2)$ → $(-2, 1)$
- $(10, 3)$ → $(3, 10)$

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

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Part 2: Finding Inverses

We are given equations and asked to find the inverse function. To do this:

1. Replace $y$ with $x$ and $x$ with $y$.
2. Solve for $y$.

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

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#### 7. $y = 2x - 5$

Swap:
$x = 2y - 5$

Solve:
$x + 5 = 2y$
$y = \frac{x + 5}{2}$

✔ Inverse: $y = \frac{x + 5}{2}$

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#### 8. $y = -3x + 5$

Swap:
$x = -3y + 5$

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

✔ Inverse: $y = \frac{5 - x}{3}$

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#### 9. $y = -5x + 10$

Swap:
$x = -5y + 10$

Solve:
$x - 10 = -5y$
$y = \frac{10 - x}{5}$

✔ Inverse: $y = \frac{10 - x}{5}$

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#### 10. $y = \frac{3}{8}x + 8$

Swap:
$x = \frac{3}{8}y + 8$

Solve:
$x - 8 = \frac{3}{8}y$
Multiply both sides by 8:
$8(x - 8) = 3y$
$8x - 64 = 3y$
$y = \frac{8x - 64}{3}$

✔ Inverse: $y = \frac{8x - 64}{3}$

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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 can't define an inverse as a function over all real numbers.

However, if we assume we want the inverse relation:

Swap:
$x = y^2 + 5$

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

⚠️ This is not a function (fails vertical line test), but the inverse relation is $y = \pm \sqrt{x - 5}$, $x \geq 5$

✔ Inverse Relation: $y = \pm \sqrt{x - 5}$, $x \geq 5$

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

Same issue: quadratic.

Swap:
$x = y^2 - 4$

Solve:
$y^2 = x + 4$
$y = \pm \sqrt{x + 4}$, $x \geq -4$

✔ Inverse Relation: $y = \pm \sqrt{x + 4}$, $x \geq -4$

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

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

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

So inverse relation: $y = -3 \pm \sqrt{x}$, $x \geq 0$

✔ Inverse Relation: $y = -3 \pm \sqrt{x}$, $x \geq 0$

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

Swap:
$x = (y - 6)^2$

Solve:
$y - 6 = \pm \sqrt{x}$
$y = 6 \pm \sqrt{x}$

✔ Inverse Relation: $y = 6 \pm \sqrt{x}$, $x \geq 0$

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

Given domain: $x \geq 2$, range: $y \geq 0$

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

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

But now we must consider the original range of $y$ was $[0, \infty)$, so the domain of the inverse is $x \geq 0$

✔ 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:
$x^2 = y + 5$
$y = x^2 - 5$

Domain of inverse: $x \geq 0$

✔ Inverse: $y = x^2 - 5$, $x \geq 0$

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

Original: $y = \sqrt{x} + 8$, so $x \geq 0$, $y \geq 8$

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

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

Now, 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:
$x + 7 = \sqrt{y}$
Square:
$(x + 7)^2 = y$

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

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

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Part 3: Verifying Inverses

Two functions $f$ and $g$ are inverses if:
- $f(g(x)) = x$
- $g(f(x)) = x$

We'll check both compositions.

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

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

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

Note: The problem says $g(x) = -\frac{1}{3}x - 3$

Check:

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

✔ Yes, inverses

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#### 22. $f(x) = 2x - 7$, $g(x) = x + 7$

Wait — $g(x) = x + 7$?

Check:

- $f(g(x)) = f(x + 7) = 2(x + 7) - 7 = 2x + 14 - 7 = 2x + 7 \ne x$

✘ Not equal to $x$. So not inverses.

But let’s see what the inverse of $f(x) = 2x - 7$ should be:

Solve $y = 2x - 7$ → $x = \frac{y + 7}{2}$ → $g(x) = \frac{x + 7}{2}$

But here $g(x) = x + 7$, which is wrong.

✔ No, not inverses

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#### 23. $f(x) = -4x + 8$, $g(x) = -\frac{3}{4}x + 2$

Wait — $g(x) = -\frac{3}{4}x + 2$? That seems odd.

Let’s compute:

- $f(g(x)) = f\left(-\frac{3}{4}x + 2\right) = -4\left(-\frac{3}{4}x + 2\right) + 8 = 3x - 8 + 8 = 3x \ne x$

Not good.

What should the inverse be?

$f(x) = -4x + 8$

Swap: $x = -4y + 8$ → $4y = 8 - x$ → $y = \frac{8 - x}{4} = 2 - \frac{x}{4}$

So inverse should be $g(x) = -\frac{1}{4}x + 2$

But given $g(x) = -\frac{3}{4}x + 2$ → different.

So ✘ Not inverses

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

Check:

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

✔ Yes, inverses

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✔ Final Answers Summary

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Inverse Relations

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

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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. $y = \frac{x + 5}{2}$
8. $y = \frac{5 - x}{3}$
9. $y = \frac{10 - x}{5}$
10. $y = \frac{8x - 64}{3}$
11. $y = \pm \sqrt{x - 5}, x \geq 5$
12. $y = \pm \sqrt{x + 4}, x \geq -4$
13. $y = -3 \pm \sqrt{x}, x \geq 0$
14. $y = 6 \pm \sqrt{x}, x \geq 0$
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. $y = (x + 7)^2, x \geq -7$

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Verifying Inverses

19. ✔ Yes
20. ✔ Yes
21. ✔ Yes
22. ✘ No
23. ✘ No
24. ✔ Yes

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