Algebra 2 - Inverses of Functions 9/10/12 - Room 148 ~~~ Pam Wilson - Free Printable

Let's solve each of these problems by finding the inverse of the given function. To find the inverse of a function, we follow these steps:

Steps to Find the Inverse:

1. Start with the equation $ y = f(x) $.
2. Swap $ x $ and $ y $: $ x = f(y) $.
3. Solve for $ y $ in terms of $ x $.
4. The resulting expression is the inverse function: $ y = f^{-1}(x) $.

We'll go through each one carefully.

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(1) $ y = 3x - 12 $

Swap $ x $ and $ y $:
$$
x = 3y - 12
$$

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

✔ Inverse: $ y = \frac{x + 12}{3} $

---

(2) $ y = -5x + 10 $

Swap $ x $ and $ y $:
$$
x = -5y + 10
$$

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

✔ Inverse: $ y = \frac{10 - x}{5} $

---

(3) $ y = 7x - 14 $

Swap $ x $ and $ y $:
$$
x = 7y - 14
$$

Solve for $ y $:
$$
x + 14 = 7y \\
y = \frac{x + 14}{7}
$$

✔ Inverse: $ y = \frac{x + 14}{7} $

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(4) $ y = 8x^2 - 9 $

This is a quadratic function. Note: Quadratic functions are not one-to-one over their entire domain, so they do not have inverses unless we restrict the domain.

But let’s proceed formally (assuming we want an inverse relation):

Swap $ x $ and $ y $:
$$
x = 8y^2 - 9
$$

Solve for $ y $:
$$
x + 9 = 8y^2 \\
y^2 = \frac{x + 9}{8} \\
y = \pm \sqrt{\frac{x + 9}{8}}
$$

So the inverse is not a function unless we restrict the domain (e.g., only take positive or negative root).

✔ Inverse relation: $ y = \pm \sqrt{\frac{x + 9}{8}} $

> ⚠️ Note: This is not a function unless domain is restricted.

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(5) $ y = 4x^2 + 7 $

Again, quadratic → not one-to-one.

Swap $ x $ and $ y $:
$$
x = 4y^2 + 7
$$

Solve for $ y $:
$$
x - 7 = 4y^2 \\
y^2 = \frac{x - 7}{4} \\
y = \pm \sqrt{\frac{x - 7}{4}} = \pm \frac{\sqrt{x - 7}}{2}
$$

✔ Inverse relation: $ y = \pm \frac{\sqrt{x - 7}}{2} $

> Again, not a function unless domain is restricted.

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(6) $ y = 11x^2 - 13 $

Same as above — quadratic.

Swap $ x $ and $ y $:
$$
x = 11y^2 - 13
$$

Solve for $ y $:
$$
x + 13 = 11y^2 \\
y^2 = \frac{x + 13}{11} \\
y = \pm \sqrt{\frac{x + 13}{11}}
$$

✔ Inverse relation: $ y = \pm \sqrt{\frac{x + 13}{11}} $

> Not a function without domain restriction.

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(7) $ y = \sqrt{x} + 5 $

Note: $ \sqrt{x} $ is defined for $ x \geq 0 $. Also, this function is one-to-one on its domain.

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

Solve for $ y $:
$$
x - 5 = \sqrt{y}
$$

Now square both sides:
$$
(x - 5)^2 = y
$$

But we must consider the domain and range.

Original function:
- Domain: $ x \geq 0 $
- Range: $ y \geq 5 $

So the inverse will have:
- Domain: $ x \geq 5 $
- Range: $ y \geq 0 $

Thus:
$$
y = (x - 5)^2, \quad \text{with } x \geq 5
$$

✔ Inverse: $ y = (x - 5)^2 $, $ x \geq 5 $

> But if we just write the expression: $ y = (x - 5)^2 $, it's technically the inverse only if we restrict the domain appropriately.

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✔ Final Answers:

| Problem | Function | Inverse |
|--------|---------|--------|
| (1) | $ y = 3x - 12 $ | $ y = \frac{x + 12}{3} $ |
| (2) | $ y = -5x + 10 $ | $ y = \frac{10 - x}{5} $ |
| (3) | $ y = 7x - 14 $ | $ y = \frac{x + 14}{7} $ |
| (4) | $ y = 8x^2 - 9 $ | $ y = \pm \sqrt{\frac{x + 9}{8}} $ |
| (5) | $ y = 4x^2 + 7 $ | $ y = \pm \frac{\sqrt{x - 7}}{2} $ |
| (6) | $ y = 11x^2 - 13 $ | $ y = \pm \sqrt{\frac{x + 13}{11}} $ |
| (7) | $ y = \sqrt{x} + 5 $ | $ y = (x - 5)^2 $, $ x \geq 5 $ |

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🔍 Summary Notes:

- Linear functions (1–3): Have inverses that are also linear.
- Quadratic functions (4–6): Do not have inverses unless domain is restricted.
- Square root function (7): Has an inverse, but domain must be considered.

Let me know if you'd like the graphs or domain restrictions explained further!