3. Define f : [−1, 1] → (−4, 4) by f(x) = 1 − 2x.

(a). Find the range of f.

(b). Does f have an inverse function? Explain.

(c). Does f have an inverse relation? Explain.

Let's analyze the function \( f \) defined as \( f(x) = 1 - 2x \) on the interval \( [-1, 1] \).

### (a) Find the range of \( f \).

To find the range of \( f \), we need to determine the output values of \( f(x) \) when \( x \) varies over the interval \([-1, 1]\).

Given \( f(x) = 1 - 2x \):

- When \( x = -1 \):
  \[
  f(-1) = 1 - 2(-1) = 1 + 2 = 3
  \]

- When \( x = 1 \):
  \[
  f(1) = 1 - 2(1) = 1 - 2 = -1
  \]

Since \( f(x) \) is a linear function (which is continuous and monotonic) and \( x \) ranges from \(-1\) to \( 1 \), \( f(x) \) will take all values between \( f(-1) = 3 \) and \( f(1) = -1 \).

Thus, the range of \( f \) is:
\[
\text{Range of } f = [-1, 3]
\]

### (b) Does \( f \) have an inverse function? Explain.

A function has an inverse if it is bijective, i.e., it must be both injective (one-to-one) and surjective (onto).

- **Injective**: A function \( f \) is injective if \( f(a) = f(b) \) implies \( a = b \).

  For \( f(x) = 1 - 2x \):
  \[
  f(a) = f(b) \implies 1 - 2a = 1 - 2b \implies -2a = -2b \implies a = b
  \]
  This shows that \( f \) is injective.

- **Surjective**: A function \( f \) from domain \( D \) to codomain \( C \) is surjective if for every element \( y \in C \), there exists an \( x \in D \) such that \( f(x) = y \).

  Given that \( f(x) \) maps \( [-1, 1] \) to \([-1, 3]\), the function \( f \) is surjective if its codomain is \([-1, 3]\). Since the given codomain is \((-4, 4)\), which includes \([-1, 3]\), \( f \) is surjective onto \([-1, 3]\).

Since \( f \) is both injective and surjective onto \([-1, 3]\), \( f \) is bijective and hence has an inverse function when considered on \([-1, 1] \to [-1, 3]\).

### (c) Does \( f \) have an inverse relation? Explain.

An inverse relation exists if for every \( y \) in the range of \( f \), there is an \( x \) in the domain of \( f \) such that \( f(x) = y \).

Since \( f(x) \) is a continuous, monotonic (strictly decreasing) function on \([-1, 1]\), for every \( y \in [-1, 3] \), there exists a unique \( x \in [-1, 1] \) such that \( f(x) = y \).

Thus, the inverse relation \( f^{-1} \) exists and it coincides with the inverse function because \( f \) is bijective on the specified domain and range.

Would you like more details or have any questions on this topic?

Here are 8 related questions for further exploration:
1. What is the inverse function of \( f(x) = 1 - 2x \)?
2. How do you determine if a function is injective graphically?
3. How do you determine if a function is surjective graphically?
4. Can a function be bijective if it is only injective but not surjective?
5. How does the continuity of a function affect its invertibility?
6. What is the significance of the domain and range in defining the inverse function?
7. How would the inverse relation of \( f(x) = 1 - 2x \) be represented graphically?
8. If the function were \( f(x) = 1 - 2x \) but defined on a different interval, would the inverse still exist?

**Tip:** When checking if a function has an inverse, always verify both injectivity and surjectivity. A function must satisfy both conditions to be invertible.

Explore the function f(x) = 1 - 2x defined on the interval [-1, 1]. Discover its range, inverse function existence, and inverse relation. Suitable for high school students studying functions and their properties.

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