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Let's work through the problems step by step.

### A. Determine whether the following functions are the same or not:
- \( f(x) = \sqrt{x} + 1 + \sqrt{x-2} \)
- \( g(x) = \sqrt{(x+1)(x-2)} \)

**Solution:**

#### 1. \( f(x) = \sqrt{x} + 1 + \sqrt{x - 2} \):
For \( f(x) \), we have two square roots and a constant added:
\[
f(x) = \sqrt{x} + 1 + \sqrt{x-2}
\]
This function is only defined for \( x \geq 2 \), because \( \sqrt{x-2} \) is undefined for \( x < 2 \).

#### 2. \( g(x) = \sqrt{(x+1)(x-2)} \):
Here, \( g(x) \) involves a square root of a product of two terms:
\[
g(x) = \sqrt{(x+1)(x-2)}
\]
We can expand \( (x+1)(x-2) \) as:
\[
(x+1)(x-2) = x^2 - 2x + x - 2 = x^2 - x - 2
\]
Thus, \( g(x) = \sqrt{x^2 - x - 2} \).

Since \( \sqrt{x^2 - x - 2} \) is not equivalent to the form of \( f(x) \) after simplification, the two functions are **not the same**.

---

### B. Determine whether the following functions are one-to-one or not:

#### i) \( f(x) = \sqrt{1 - x^2}, \; 0 \leq x \leq 1 \)
This is the upper half of a semicircle with radius 1. Since it's restricted to \( 0 \leq x \leq 1 \), the function is **not one-to-one**, as the function does not have a unique output for each input over a larger domain, but within this domain it is fine.

#### ii) \( f(x) = 3 - (5 + x)^{1/7} \)
This is a decreasing function because of the negative exponent. As it is strictly decreasing, the function is **one-to-one**.

#### iii) \( f(x) = \cos(x) \)
The cosine function is periodic and repeats its values over its domain, so it is **not one-to-one**.

---

### C. Assume that \( f(x) = \sqrt[3]{1 - x^3} \) is a one-to-one function:

#### i) Find \( f^{-1} \):
To find the inverse, start by setting \( y = f(x) \):
\[
y = \sqrt[3]{1 - x^3}
\]
Now, cube both sides:
\[
y^3 = 1 - x^3
\]
Solving for \( x^3 \):
\[
x^3 = 1 - y^3
\]
Now take the cube root:
\[
x = \sqrt[3]{1 - y^3}
\]
Thus, the inverse function is:
\[
f^{-1}(x) = \sqrt[3]{1 - x^3}
\]

#### ii) Domain and range of \( f^{-1} \):
- The domain of \( f(x) \) (original function) is all real numbers because cube roots are defined for all real values.
- The range of \( f(x) \) is also all real numbers because \( \sqrt[3]{1 - x^3} \) can take any real value.

Therefore, the domain and range of \( f^{-1}(x) \) are both all real numbers \( (-\infty, \infty) \).

---

### D. Write the function \( h(x) = \frac{1}{(x - 2)^3} \) as a composition of two functions.

Let:
- \( u(x) = x - 2 \)
- \( v(u) = \frac{1}{u^3} \)

Thus, we can write:
\[
h(x) = v(u(x)) = \frac{1}{(x - 2)^3}
\]
So the function can be written as:
\[
h(x) = v(u(x))
\]

---

If you would like any further explanations, feel free to ask!

Here are 5 related questions for practice:
1. How do you verify if two functions are equal?
2. How can you determine if a function is one-to-one by using its derivative?
3. What is the general method for finding the inverse of a function?
4. What is the difference between the domain and range of a function and its inverse?
5. How can you decompose a more complex function into simpler functions?

**Tip:** Always check the domain restrictions when determining if two functions are equal!

The image contains four math problems related to function equivalence, one-to-one functions, inverse functions, and composition of functions.

This page provides step-by-step solutions to function comparison, determining one-to-one functions, finding the inverse of f(x), and writing functions as compositions. It is suitable for advanced high school or college students, focusing on key concepts like inverse functions and function composition.

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