Proving The Range of a Function

# Proving The Range of a Function

We are now going to learn how to rigorously prove the range of a function is a certain set. We will use the notation, "$R(f)$" to denote the set that is the range of $f$. Before we look at some examples, be sure to recall from the Proving Set Theorems page that to show two sets $A$ and $B$ are equal, that is $A = B$ , we must show that $A \subseteq B$ and $B \subseteq A$.

## Example 1

For $x \in \mathbb{R}$ let $f(x) = \frac{-3x}{\sqrt{4 + x^2}}$ be a function. Prove that $R(f) = (-3, 3)$.

Step 1: We will first show that $R(f) \subseteq (-3, 3)$. Let $y \in R(f)$ so that $y = f(x)$ for some $x \in \mathbb{R}$. We note that $y = 0$ if and only if $x = 0$. Now if $y \neq 0$, then we have that $4 + x^2 > x^2 > 0$, and so $\sqrt{4 + x^2}$ and $\sqrt{x^2}$ are defined, and it follows that:

(1)
\begin{align} \quad \mid y \mid = \mid f(x) \mid = \frac{3\mid x \mid}{\sqrt{4 + x^2}} < \frac{3 \mid x \mid}{\sqrt{x^2}} = \frac{3 \mid x \mid}{\mid x \mid} = 3 \end{align}

Therefore $\mid y \mid < 3$ and so $-3 < y < 3$ so $y \in (-3, 3)$.

Step 2: We now want to show that $(-3, 3) \subseteq R(f)$. Let $y \in (-3, 3)$ and so $-3 < y < 3$ which implies that $0 ≤ y^2 < 9$ and so $9 - y^2 > 0$. Doing some algebraic manipulations on $f$ we have that:

(2)
\begin{align} y = \frac{-3x}{\sqrt{4 + x^2}} \\ y \sqrt{4 + x^2} = -3x \\ y^2 (4 + x^2) = 9x^2 \\ 4y^2 + y^2x^2 = 9x^2 \\ 4y^2 = 9x^2 - y^2x^2 \\ 4y^2 = x^2(9 - y^2) \\ x^2 = \frac{4y^2}{9 - y^2} \\ \sqrt{x^2} = \sqrt{\frac{4y^2}{9 - y^2}} \\ \mid x \mid = \frac{\sqrt{4y^2}}{\sqrt{9 - y^2}} \\ x = \frac{-2y}{\sqrt{9 - y^2}} \end{align}

So there exists an $x \in \mathbb{R}$ such that $y = f(x)$ and so $y \in R(f)$. Therefore $(-3, 3) \subseteq R(f)$.

Since $R(f) \subseteq (-3, 3)$ and $(-3, 3) \subseteq R(f)$ we have that $R(f) = (-3, 3)$.

## Example 2

For $x \in \mathbb{R}$ let $f(x) = \frac{-2x}{\sqrt{1 + x^2}}$ be a function. Prove that $R(f) = (-2, 2)$.

Step 1: We will first show that $R(f) \subseteq (-2, 2)$. Let $y \in R(f)$ so that $y = f(x)$ for some $x \in \mathbb{R}$. We first note that $y = 0$ if and only if $x = 0$. Now if $y \neq 0$, then $1 + x^2 > x^2 > 0$. Therefore $\sqrt{1 + x^2}$ and $\sqrt{x^2}$ are defined, and so it follows that:

(3)
\begin{align} \quad \mid y \mid = \mid f(x) \mid = \frac{\mid -2x \mid}{\sqrt{x^2}} < \frac{2 \mid x \mid}{\mid x \mid} = 2 \end{align}

Therefore $\mid y \mid < 2$ and we have that $-2 < y < 2$ and so $y \in (-2, 2)$. Therefore $R(f) \subseteq (-2, 2)$.

Step 2: We now want to show that $(-2, 2) \subseteq R(f)$. Let $y \in (-2, 2)$. We need to show that there exists an $x \in \mathbb{R}$ such that $y = f(x)$. We will first so some algebraic manipulation of our function $f$ as follows:

(4)
\begin{align} y = \frac{-2x}{\sqrt{1 + x^2}} \\ y \sqrt{1 + x^2} = -2x \\ y^2 (1 + x^2) = 4x^2 \\ y^2 + y^2x^2 = 4x^2 \\ y^2 = 4x^2 - y^2x^2 \\ y^2 = x^2(4 - y^2) \\ x^2 = \frac{y^2}{4 - y^2} \\ \sqrt{x^2} = \sqrt{\frac{y^2}{4 - y^2}} \\ \mid x \mid = \frac{\mid y^2 \mid}{\sqrt{4 - y^2}} \\ x = \frac{-y}{\sqrt{4 - y^2}} \end{align}

Now since $y \in (-2, 2)$, then $-2 < y < 2$ and so $0 ≤ y^2 < 4$, which is equivalent to saying $4 - y^2 > 0$. So there exists an $x \in \mathbb{R}$ such that $y = f(x)$ and so $y \in R(f)$. Therefore $(-2, 2) \subseteq R(f)$.

Since $R(f) \subseteq (-2, 2)$ and $(-2, 2) \subseteq R(f)$ we have that $R(f) = (-2, 2)$.

## Example 3

For $x \in \mathbb{R}$ let $f(x) = x^2$. Prove that $R(f) = [0, \infty)$.

Step 1: We will first show that $R(f) \subseteq [0, \infty)$. Let $y \in [0, \infty)$ so that $y = f(x)$ for some $x \in \mathbb{R}$. Note that $x^2 > 0$ for all $x \in \mathbb{R}$, and since $y \in [0, \infty)$, then $y \in [0, \infty)$ and so $R(f) \subseteq [0, \infty)$.

Step 2: We now want to show that $[0, \infty) \subseteq R(f)$. Let $y \in [0, \infty)$. Therefore $0 ≤ y < \infty$ and so $\sqrt{y}$ is well defined. We need to show that for some $x \in \mathbb{R}$, $y = f(x)$. Notice that:

(5)
\begin{align} y = x^2 \\ \sqrt{y} = x \end{align}

So there exists an $x$ such that $y = f(x)$, and so $[0, \infty) \subseteq R(f)$.

Since $R(f) \subseteq [0, \infty)$ and $[0, \infty) \subseteq R(f)$ we have that $R(f) = [0, \infty)$.