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!