Example

What is the domain and range of the function \(f(x)=4-x^2\)?

Solution

Here a graph of the function helps.

Since \(f(x)\) is defined for all real numbers, we have \(\mathrm{domain}(f) = \mathbb{R}\). We can see from the graph that \(\mathrm{range}(f) = \{\, y : y\leq 4 \,\} = (-\infty,4]\).

Example

What is the domain and range of \(f(x)=-\dfrac{7}{x}\)?

Solution

The graph of \(y=f(x)\) is a rectangular hyperbola.

From the graph, we can see that \begin{align*} \mathrm{domain}(f) &= \{\, x\in\mathbb{R} \mid x\neq 0 \,\} = \mathbb{R}\setminus \{0\} \\ \mathrm{range}(f) &= \{\, y\in\mathbb{R} \mid y\neq 0 \,\} = \mathbb{R}\setminus \{0\}. \end{align*}

Example

What is the domain and range of \(f(x)=7^x-5\)?

Solution

This is a vertical translation of an exponential function.

Hence, \(\mathrm{domain}(f) = \mathbb{R}\) and \(\mathrm{range}(f) = (-5,\infty)\).

Example

What is the domain and range of \(f(x)=\sqrt{16-x^2}\)?

Solution

We have \begin{align*} y = \sqrt{16-x^2} \ &\iff\ y^2 = 16-x^2 \hspace{4mm}\text{and}\hspace{4mm} y \geq 0 \\ &\iff\ x^2+y^2=16 \hspace{4mm}\text{and}\hspace{4mm} y \geq 0. \end{align*}

So the graph of \(f(x)\) is the top half of the circle with centre the origin and radius 4.

Hence, \(\mathrm{domain}(f) = [-4,4]\) and \(\mathrm{range}(f) = [0,4]\).

Example

Find the domain and range of \(f(x)=\sqrt{1-2x}\).

Solution

We first find the domain of \(f(x)\):

\begin{align*} f(x) \text{ is defined} \ &\iff\ 1-2x \geq 0 \\ &\iff\ x \leq \tfrac{1}{2}. \end{align*}

So \(\mathrm{domain}(f) = (-\infty,\dfrac{1}{2}]\).

Every non-negative real number can be obtained as a value of \(f(x) = \sqrt{1-2x}\). Thus \(\mathrm{range}(f) = [0,\infty)\).

We can sketch the graph of this function by noticing that

\begin{align*} y = \sqrt{1-2x} \ &\iff\ y^2 = 1-2x \hspace{4mm}\text{and}\hspace{4mm} y \geq 0 \\ &\iff\ x = \tfrac{1}{2}(1 - y^2) \hspace{4mm}\text{and}\hspace{4mm} y \geq 0. \end{align*}

So this is 'half' of a rotated and translated parabola.

Example

Find the domain and range of \(f(x)=x^{\frac{1}{3}}\). Sketch the graph of \(y=f(x)\).

Solution

Clearly \(\mathrm{domain}(f) = \mathbb{R}\), since all real numbers have a cube root. As \(x\to \infty\), \(f(x)\to\infty\) and as \(x\to -\infty\), \(f(x)\to -\infty\). Since \(f\) is continuous, it follows that \(\mathrm{range}(f) = \mathbb{R}\).

We have \(y=x^{\frac{1}{3}}\) if and only if \(x=y^3\). The graph of \(y=x^3\) is as follows.

The graph of \(x=y^3\), or \(y=x^\frac{1}{3}\), is obtained by reflecting in the line \(y = x\).

Example

Find the domain and range of \(f(x)=|x|\), where as usual

\[ |x| = \begin{cases} -x &\text{if } x < 0, \\ x &\text{if } x \geq 0. \end{cases} \]

Solution

By drawing the graph of \(y=|x|\), we see that \(\mathrm{domain}(f) = \mathbb{R}\) and \(\mathrm{range}(f) = [0,\infty)\).