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### Question 3: Part (a)

Determine whether \( f \) is a function from \( \mathbb{R} \to \mathbb{R} \) for each given \( f(x) \). If \( f \) is not a function, provide a reason.

#### (i) \( f(x) = \frac{1}{x^2 - 9} \)

- This function is not defined when \( x^2 - 9 = 0 \), or \( x = \pm 3 \). 
- Hence, \( f(x) \) is **not** a function from \( \mathbb{R} \to \mathbb{R} \) because it is undefined at \( x = 3 \) and \( x = -3 \).

#### (ii) \( f(x) = \sqrt{2 + x^2} \)

- The square root function is only defined for non-negative numbers. Since \( x^2 \) is always non-negative, \( 2 + x^2 \geq 2 \), so the square root is always defined.
- Therefore, \( f(x) \) **is** a function from \( \mathbb{R} \to \mathbb{R} \).

#### (iii) \( f(x) = x^2 \pm 2 \)

- The notation \( \pm \) suggests two possible values for \( f(x) \), leading to an ambiguity.
- Since each \( x \) can have two values for \( f(x) \), this means \( f \) is not a function (because a function must assign exactly one value to each \( x \)).
- Hence, \( f(x) \) is **not** a function.

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### Question 3: Part (b)

The functions \( f: \mathbb{R} \to \mathbb{R} \), \( g: \mathbb{R} \to \mathbb{R} \), and \( h: \mathbb{R} \to \mathbb{R} \) are given by:
- \( f(x) = x^2 - 1 \)
- \( g(x) = 2 - 3x \)
- \( h(x) = \frac{2x - 5}{3} \)

#### (i) Is \( h \) a one-to-one function? Prove or give a counterexample.

- A function is one-to-one if different inputs lead to different outputs. For \( h(x) = \frac{2x - 5}{3} \), let’s check if it is injective (one-to-one).
- Suppose \( h(x_1) = h(x_2) \). This gives:
  \[
  \frac{2x_1 - 5}{3} = \frac{2x_2 - 5}{3}
  \]
  \[
  2x_1 - 5 = 2x_2 - 5
  \]
  \[
  x_1 = x_2
  \]
- Since \( x_1 = x_2 \) implies that the inputs must be the same, \( h(x) \) is one-to-one.

#### (ii) Find the inverse of \( g(x) \).

The function \( g(x) = 2 - 3x \). To find the inverse, solve for \( x \) in terms of \( y \):
- \( y = 2 - 3x \)
- \( 3x = 2 - y \)
- \( x = \frac{2 - y}{3} \)

Thus, the inverse function \( g^{-1}(x) = \frac{2 - x}{3} \).

#### (iii) Find the composition \( (g \circ f)(x) \). Express the solution in its simplest form.

The composition \( (g \circ f)(x) = g(f(x)) \):
- \( f(x) = x^2 - 1 \), so:
  \[
  g(f(x)) = g(x^2 - 1) = 2 - 3(x^2 - 1) = 2 - 3x^2 + 3 = 5 - 3x^2
  \]
Thus, \( (g \circ f)(x) = 5 - 3x^2 \).

#### (iv) Given that \( (f \circ m)(x) = 4x(x + 1) \), find the function \( m(x) \).

- \( f(x) = x^2 - 1 \), and \( (f \circ m)(x) = f(m(x)) \).
- We need to find \( m(x) \) such that:
  \[
  f(m(x)) = 4x(x + 1)
  \]
- First, express \( 4x(x + 1) \) in expanded form:
  \[
  4x(x + 1) = 4x^2 + 4x
  \]
- Now, recall \( f(x) = x^2 - 1 \), so \( f(m(x)) = m(x)^2 - 1 \). Thus, we need:
  \[
  m(x)^2 - 1 = 4x^2 + 4x
  \]
  \[
  m(x)^2 = 4x^2 + 4x + 1
  \]
  \[
  m(x) = \pm(2x + 1)
  \]
Thus, \( m(x) = 2x + 1 \) (choosing the positive branch for simplicity).

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Would you like more details or have any questions about these solutions?

### 5 Related Questions:
1. What is the definition of a function from \( \mathbb{R} \to \mathbb{R} \)?
2. How can you determine if a function is one-to-one or injective?
3. What are the steps to finding the inverse of a linear function?
4. How does function composition work, and what are common applications of it?
5. Why does a function need to be injective to have an inverse?

### Tip:
When dealing with composite functions, always carefully track how the inner function affects the domain and range of the outer function. This helps in simplifying the expressions and verifying if the composition is valid over all real numbers.

Determine whether f is a function from R to R in the following equations. If f is not a function, give a reason. Functions provided: (i) f(x) = 1 / (x^2 - 9), (ii) f(x) = sqrt(2 + x^2), (iii) f(x) = x^2 ± 2. Additional questions related to the functions f, g(x) = 2 - 3x, and h(x) = (2x - 5) / 3 regarding injectivity, inverse, composition, and finding a missing function.

Solve problems related to determining if functions f(x), g(x), and h(x) are injective, find inverses, and handle function compositions. This is a step-by-step guide through advanced high school-level algebra, suitable for Grades 10-12.

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