Math Exercises & Math Problems: Inverse Function - Free Printable

It looks like you've uploaded a list of 26 mathematical functions labeled from a) to z), but no specific task or question was stated. Since you mentioned "solve the problem," I’ll assume that your goal is to simplify each function or determine its domain, as these are common tasks for such lists.

However, without a clear instruction (e.g., "find the domain," "simplify," "differentiate," etc.), I'll proceed with the most likely interpretation: Simplify each expression and determine its domain, where applicable.

Let’s go through them one by one and simplify or analyze them.

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a) $ y = \frac{x}{5} $

- Simplified: Already simple.
- Domain: All real numbers, $ x \in \mathbb{R} $

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b) $ y = 1 - [4x - 7 - (1 - 2x) + 3] - x $

Simplify step by step:

1. Inside the brackets:
$$
4x - 7 - (1 - 2x) + 3 = 4x - 7 - 1 + 2x + 3 = (4x + 2x) + (-7 -1 + 3) = 6x - 5
$$
2. Now:
$$
y = 1 - (6x - 5) - x = 1 - 6x + 5 - x = (1 + 5) - 7x = 6 - 7x
$$

- Simplified: $ y = 6 - 7x $
- Domain: $ \mathbb{R} $

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c) $ y = \frac{2x + 3}{3x + 5} $

- Simplified: Already simplified.
- Domain: Denominator ≠ 0 → $ 3x + 5 \neq 0 $ → $ x \neq -\frac{5}{3} $
- Domain: $ x \in \mathbb{R}, x \neq -\frac{5}{3} $

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d) $ y = \frac{x - 1}{6x + 3} $

Factor denominator: $ 6x + 3 = 3(2x + 1) $

So,
$$
y = \frac{x - 1}{3(2x + 1)}
$$

- Simplified: $ \frac{x - 1}{3(2x + 1)} $
- Domain: $ 2x + 1 \neq 0 $ → $ x \neq -\frac{1}{2} $

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e) $ y = 1 - \frac{1}{2x} $

- Simplified: Already simplified.
- Domain: $ 2x \neq 0 $ → $ x \neq 0 $

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f) $ y = \frac{-x - 7}{x + 5} $

Factor numerator: $ -x - 7 = -(x + 7) $

So,
$$
y = \frac{-(x + 7)}{x + 5}
$$

- Simplified: $ y = -\frac{x + 7}{x + 5} $
- Domain: $ x \neq -5 $

---

g) $ y = \frac{9 - 3x}{9x - 3} $

Factor numerator and denominator:
- Numerator: $ 9 - 3x = 3(3 - x) $
- Denominator: $ 9x - 3 = 3(3x - 1) $

So,
$$
y = \frac{3(3 - x)}{3(3x - 1)} = \frac{3 - x}{3x - 1}
$$

Note: $ 3 - x = -(x - 3) $, so alternatively $ y = -\frac{x - 3}{3x - 1} $

- Simplified: $ \frac{3 - x}{3x - 1} $
- Domain: $ 3x - 1 \neq 0 $ → $ x \neq \frac{1}{3} $

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h) $ y = \frac{10x - 5}{15 - 10x} + 1 $

Factor:
- Numerator: $ 10x - 5 = 5(2x - 1) $
- Denominator: $ 15 - 10x = 5(3 - 2x) = -5(2x - 3) $

So:
$$
y = \frac{5(2x - 1)}{-5(2x - 3)} + 1 = -\frac{2x - 1}{2x - 3} + 1
$$

Now combine:
$$
y = 1 - \frac{2x - 1}{2x - 3}
= \frac{(2x - 3) - (2x - 1)}{2x - 3} = \frac{2x - 3 - 2x + 1}{2x - 3} = \frac{-2}{2x - 3}
$$

- Simplified: $ y = \frac{-2}{2x - 3} $
- Domain: $ 2x - 3 \neq 0 $ → $ x \neq \frac{3}{2} $

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i) $ y = \frac{1 - [10 - (7 - x) + 20] - 5x}{1 + 2x - (3 - 4x)} - 2 $

Simplify numerator and denominator separately.

Numerator:
$$
1 - [10 - (7 - x) + 20] - 5x = 1 - [10 - 7 + x + 20] - 5x = 1 - [23 + x] - 5x = 1 - 23 - x - 5x = -22 - 6x
$$

Denominator:
$$
1 + 2x - (3 - 4x) = 1 + 2x - 3 + 4x = (1 - 3) + (2x + 4x) = -2 + 6x
$$

So:
$$
y = \frac{-22 - 6x}{-2 + 6x} - 2 = \frac{-6x - 22}{6x - 2} - 2
$$

Factor:
- Numerator: $ -2(3x + 11) $
- Denominator: $ 2(3x - 1) $

So:
$$
y = \frac{-2(3x + 11)}{2(3x - 1)} - 2 = \frac{-(3x + 11)}{3x - 1} - 2
$$

Now combine:
$$
y = \frac{-(3x + 11) - 2(3x - 1)}{3x - 1} = \frac{-3x - 11 - 6x + 2}{3x - 1} = \frac{-9x - 9}{3x - 1} = \frac{-9(x + 1)}{3x - 1}
$$

- Simplified: $ y = \frac{-9(x + 1)}{3x - 1} $
- Domain: $ 3x - 1 \neq 0 $ → $ x \neq \frac{1}{3} $

---

j) $ y = (x^3 - 1) : x^3 $

This means division: $ y = \frac{x^3 - 1}{x^3} $

$$
y = 1 - \frac{1}{x^3}
$$

- Simplified: $ y = 1 - \frac{1}{x^3} $
- Domain: $ x \neq 0 $

---

k) $ y = -x^2 - (-x)^2 $

Note: $ (-x)^2 = x^2 $

So:
$$
y = -x^2 - x^2 = -2x^2
$$

- Simplified: $ y = -2x^2 $
- Domain: $ \mathbb{R} $

---

l) $ y = 2x^{4/3} - 1 $

This is a power function.

- $ x^{4/3} = (x^{1/3})^4 $ — cube root then fourth power.
- Cube root defined for all real $ x $, so domain is all reals.

- Simplified: $ y = 2x^{4/3} - 1 $
- Domain: $ \mathbb{R} $

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m) $ y = x^3 + 3x^2 + 3x + 1 $

Recognize this as:
$$
y = (x + 1)^3
$$

- Simplified: $ y = (x + 1)^3 $
- Domain: $ \mathbb{R} $

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n) $ y = \sqrt{x - 1} $

Square root requires non-negative argument:
- $ x - 1 \geq 0 $ → $ x \geq 1 $

- Domain: $ [1, \infty) $

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o) $ y = \sqrt{(2x + 4)^3 - 7} $

For square root to be defined:
- $ (2x + 4)^3 - 7 \geq 0 $
- $ (2x + 4)^3 \geq 7 $
- $ 2x + 4 \geq \sqrt[3]{7} $
- $ 2x \geq \sqrt[3]{7} - 4 $
- $ x \geq \frac{\sqrt[3]{7} - 4}{2} $

- Domain: $ x \geq \frac{\sqrt[3]{7} - 4}{2} $

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p) $ y = (5 - \sqrt{x + 2})^4 + 3 $

First, $ \sqrt{x + 2} $ requires:
- $ x + 2 \geq 0 $ → $ x \geq -2 $

Then $ 5 - \sqrt{x + 2} $ is real for $ x \geq -2 $, and raised to 4th power is always defined.

- Domain: $ [-2, \infty) $

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q) $ y = 2^x $

Exponential function: defined for all real $ x $

- Domain: $ \mathbb{R} $

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r) $ y = \left(\frac{1}{8}\right)^{1 - x} $

Rewrite: $ \left(8^{-1}\right)^{1 - x} = 8^{-(1 - x)} = 8^{x - 1} $

Alternatively, $ y = \left(\frac{1}{8}\right)^{1 - x} $

Defined for all real $ x $

- Domain: $ \mathbb{R} $

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s) $ y = -3 \cdot 5^x + 6 $

Exponential function, defined for all $ x $

- Domain: $ \mathbb{R} $

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t) $ y = 1 + \log x $

Logarithm defined only for $ x > 0 $

- Domain: $ (0, \infty) $

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u) $ y = -2 \log\left(\frac{x - 1}{x + 1}\right)^5 $

Note: $ \log(a^5) = 5 \log a $, so:
$$
y = -2 \cdot 5 \log\left(\frac{x - 1}{x + 1}\right) = -10 \log\left(\frac{x - 1}{x + 1}\right)
$$

But we must ensure:
- $ \frac{x - 1}{x + 1} > 0 $

Sign analysis:
- Critical points: $ x = -1, 1 $
- Sign chart:
- $ x < -1 $: $ \frac{-}{-} = + $
- $ -1 < x < 1 $: $ \frac{-}{+} = - $
- $ x > 1 $: $ \frac{+}{+} = + $

So $ \frac{x - 1}{x + 1} > 0 $ when $ x < -1 $ or $ x > 1 $

Also, $ x \neq -1 $ (denominator zero)

- Domain: $ (-\infty, -1) \cup (1, \infty) $

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v) $ y = \log x - \log 2x + \log 3x $

Use log rules:
$$
y = \log\left(\frac{x \cdot 3x}{2x}\right) = \log\left(\frac{3x^2}{2x}\right) = \log\left(\frac{3x}{2}\right)
$$

But we must consider domain: all logs require positive arguments:
- $ x > 0 $
- $ 2x > 0 $ → $ x > 0 $
- $ 3x > 0 $ → $ x > 0 $

So domain: $ x > 0 $

And simplified: $ y = \log\left(\frac{3x}{2}\right) $

- Simplified: $ y = \log\left(\frac{3x}{2}\right) $
- Domain: $ (0, \infty) $

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w) $ y = \sin 2x + 1 $

Trigonometric function: defined for all real $ x $

- Domain: $ \mathbb{R} $

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x) $ y = \left(1 - \cos\frac{x}{2}\right)^2 - 1 $

No restrictions on cosine, so defined everywhere.

- Domain: $ \mathbb{R} $

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y) $ y = 2\tan^2\left(x + \frac{\pi}{2}\right) - 8 $

Recall: $ \tan\left(x + \frac{\pi}{2}\right) = -\cot x $

But better to think about domain.

$ \tan \theta $ undefined when $ \theta = \frac{\pi}{2} + k\pi $

So:
$$
x + \frac{\pi}{2} \neq \frac{\pi}{2} + k\pi \Rightarrow x \neq k\pi
$$

So $ x \neq k\pi $, $ k \in \mathbb{Z} $

- Domain: $ x \in \mathbb{R}, x \neq k\pi $

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z) $ y = \frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x} $

Numerator: $ \cos^2 x - \sin^2 x = \cos 2x $

Denominator: $ 2 \sin x \cos x = \sin 2x $

So:
$$
y = \frac{\cos 2x}{\sin 2x} = \cot 2x
$$

So $ y = \cot 2x $

Defined when $ \sin 2x \neq 0 $ → $ 2x \neq k\pi $ → $ x \neq \frac{k\pi}{2} $

- Domain: $ x \neq \frac{k\pi}{2} $, $ k \in \mathbb{Z} $

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## ✔ Summary Table

| Letter | Simplified Expression | Domain |
|--------|------------------------|--------|
| a) | $ \frac{x}{5} $ | $ \mathbb{R} $ |
| b) | $ 6 - 7x $ | $ \mathbb{R} $ |
| c) | $ \frac{2x + 3}{3x + 5} $ | $ x \neq -\frac{5}{3} $ |
| d) | $ \frac{x - 1}{3(2x + 1)} $ | $ x \neq -\frac{1}{2} $ |
| e) | $ 1 - \frac{1}{2x} $ | $ x \neq 0 $ |
| f) | $ -\frac{x + 7}{x + 5} $ | $ x \neq -5 $ |
| g) | $ \frac{3 - x}{3x - 1} $ | $ x \neq \frac{1}{3} $ |
| h) | $ \frac{-2}{2x - 3} $ | $ x \neq \frac{3}{2} $ |
| i) | $ \frac{-9(x + 1)}{3x - 1} $ | $ x \neq \frac{1}{3} $ |
| j) | $ 1 - \frac{1}{x^3} $ | $ x \neq 0 $ |
| k) | $ -2x^2 $ | $ \mathbb{R} $ |
| l) | $ 2x^{4/3} - 1 $ | $ \mathbb{R} $ |
| m) | $ (x + 1)^3 $ | $ \mathbb{R} $ |
| n) | $ \sqrt{x - 1} $ | $ [1, \infty) $ |
| o) | $ \sqrt{(2x + 4)^3 - 7} $ | $ x \geq \frac{\sqrt[3]{7} - 4}{2} $ |
| p) | $ (5 - \sqrt{x + 2})^4 + 3 $ | $ [-2, \infty) $ |
| q) | $ 2^x $ | $ \mathbb{R} $ |
| r) | $ \left(\frac{1}{8}\right)^{1 - x} $ | $ \mathbb{R} $ |
| s) | $ -3 \cdot 5^x + 6 $ | $ \mathbb{R} $ |
| t) | $ 1 + \log x $ | $ (0, \infty) $ |
| u) | $ -10 \log\left(\frac{x - 1}{x + 1}\right) $ | $ (-\infty, -1) \cup (1, \infty) $ |
| v) | $ \log\left(\frac{3x}{2}\right) $ | $ (0, \infty) $ |
| w) | $ \sin 2x + 1 $ | $ \mathbb{R} $ |
| x) | $ \left(1 - \cos\frac{x}{2}\right)^2 - 1 $ | $ \mathbb{R} $ |
| y) | $ 2\tan^2\left(x + \frac{\pi}{2}\right) - 8 $ | $ x \neq k\pi $ |
| z) | $ \cot 2x $ | $ x \neq \frac{k\pi}{2} $ |

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If you had a specific task (like finding derivatives, limits, graphing, or identifying types), please let me know! But based on the list, simplification and domain determination is the most logical interpretation.

Let me know if you'd like any further analysis (e.g., derivative, continuity, asymptotes).