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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