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 address each part of the problem for the function \( f : [-1, 1] \rightarrow (-4, 4) \) defined by \( f(x) = 1 - 2x \).

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

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

1. Calculate \( f(x) \) at the endpoints of the interval:
   - For \( x = -1 \):
     \[
     f(-1) = 1 - 2(-1) = 1 + 2 = 3
     \]
   - For \( x = 1 \):
     \[
     f(1) = 1 - 2(1) = 1 - 2 = -1
     \]

2. Since \( f(x) = 1 - 2x \) is a linear function (hence continuous and monotonous), the values of \( f(x) \) will cover all numbers between \( f(-1) \) and \( f(1) \).

Therefore, the range of \( f \) is:
\[
[-1, 3]
\]

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

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

- **Injectivity**: A function \( f \) is injective if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). For \( f(x) = 1 - 2x \):
  \[
  f(x_1) = f(x_2) \implies 1 - 2x_1 = 1 - 2x_2 \implies -2x_1 = -2x_2 \implies x_1 = x_2
  \]
  Therefore, \( f \) is injective.

- **Surjectivity**: A function \( f \) is surjective if for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). Since the codomain of \( f \) is \((-4, 4)\) but the range is \([-1, 3]\), \( f \) is not surjective onto \((-4, 4)\). Therefore, \( f \) is not surjective onto its given codomain.

Since \( f \) is not surjective onto \((-4, 4)\), \( f \) does not have an inverse function as defined from \([-1, 1]\) to \((-4, 4)\).

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

An inverse relation exists if we can express \( x \) in terms of \( y \) from \( y = f(x) \). Even if \( f \) is not bijective, we can find an inverse relation by solving for \( x \) in terms of \( y \):

Given \( y = 1 - 2x \), solve for \( x \):
\[
y = 1 - 2x \implies 2x = 1 - y \implies x = \frac{1 - y}{2}
\]

The inverse relation can be written as:
\[
f^{-1}(y) = \frac{1 - y}{2}
\]

However, this inverse relation is valid only for \( y \) within the range of \( f \), which is \([-1, 3]\).

### Summary

(a) The range of \( f \) is \([-1, 3]\).

(b) \( f \) does not have an inverse function because it is not surjective onto its codomain \((-4, 4)\).

(c) \( f \) does have an inverse relation given by \( f^{-1}(y) = \frac{1 - y}{2} \), valid for \( y \in [-1, 3] \).

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Explore the function f(x) = 1 - 2x defined on the interval [-1, 1]. Learn about its range, the conditions for having an inverse function, and the concept of inverse relations. Suitable for high school students studying functions and calculus.

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