The square root function and domains

Square roots bring a new restric­tion into func­tions: in real num­bers you can­not take the square root of a neg­a­tive num­ber. This sin­gle fact is behind a whole class of ques­tions that ask for the domain of a func­tion. In this les­son you will meet the square root func­tion and its graph, and then learn to find the domain of expres­sions such as $\sqrt{x^2 - 5x + 6}$ or $\displaystyle \frac{1}{\sqrt{x - 2}}$ by turn­ing the con­di­tion into an inequal­ity you already know how to solve.

Square roots of real num­bers#

If $a^2 = b$ with $a \geq 0$, then $\sqrt{b} = a$. For exam­ple,

$$2^2 = 4 \implies \sqrt{4} = 2, \qquad 3^2 = 9 \implies \sqrt{9} = 3$$

The sym­bol $\sqrt{\phantom{x}}$ always means the non-neg­a­tive square root. Although $(-2)^2 = 4$ too, $\sqrt{4}$ is $2$, not $-2$.

The square root func­tion#

The func­tion $y = f(x) = \sqrt{x}$ is called the square root func­tion.

Since only non-neg­a­tive num­bers have real square roots, and a square root is never neg­a­tive,

$$\text{Domain} = [0, \infty), \qquad \text{Range} = [0, \infty)$$

The graph of $y = \sqrt{x}$#

The graph starts at the ori­gin and rises to the right, pass­ing through $(1, 1)$, $(4, 2)$ and $(9, 3)$. It rises steeply at first and then more and more slowly. There is no graph at all to the left of the $y$-axis.

Domains of square root expres­sions#

Rule 1. $\sqrt{T}$ is defined (as a real num­ber) exactly when $T \geq 0$.

Rule 2. $\displaystyle \frac{1}{\sqrt{T}}$ is defined exactly when $T > 0$. The square root needs $T \geq 0$, and the denom­i­na­tor can­not be $0$, so $T = 0$ is also excluded.

To find a domain, write the con­di­tion from the rule and solve it.

Exam­ple 1#

Find the domain of $\sqrt{x - 2}$.

$\sqrt{x - 2}$ is defined when

$$x - 2 \geq 0 \implies x \geq 2 \implies x \in [2, \infty)$$

So the domain of $\sqrt{x - 2}$ is $[2, \infty)$.

Exam­ple 2#

Find the domain of $\sqrt{2x - 3}$.

$$\displaystyle 2x - 3 \geq 0 \implies x \geq \frac{3}{2} \implies x \in \left[\frac{3}{2}, \infty\right)$$

Exam­ple 3#

Find the domain of $\sqrt{x^2 - 5x + 6}$.

We need $x^2 - 5x + 6 \geq 0$, that is, $(x - 2)(x - 3) \geq 0$. This is an upward parabola with roots $2$ and $3$; it is non-neg­a­tive out­side and at the roots:

$$x \in (-\infty, 2] \cup [3, \infty)$$

Exam­ple 4#

Find the domain of $\sqrt{x - |x|}$.

We need $x - |x| \geq 0$, that is, $|x| \leq x$. But $|x| < x$ never hap­pens, so this means $|x| = x$, which is true exactly when $x \geq 0$.

So the domain is $[0, \infty)$. (For neg­a­tive $x$, $x - |x| = 2x$ is neg­a­tive.)

Exam­ple 5#

Find the domain of $\displaystyle \frac{1}{\sqrt{x - 2}}$.

By Rule 2,

$$x - 2 > 0 \implies x > 2 \implies x \in (2, \infty)$$

Com­pare with Exam­ple 1: the only dif­fer­ence is the point $x = 2$, which is allowed for $\sqrt{x - 2}$ but not for $\displaystyle \frac{1}{\sqrt{x - 2}}$.

Exam­ple 6#

Find the domain of $\sqrt{4 - x^2}$.

We need $4 - x^2 \geq 0$, that is, $x^2 - 4 \leq 0$, or $(x + 2)(x - 2) \leq 0$. Less than zero, between the roots:

$$x \in [-2, 2]$$

Exam­ple 7#

Find the domain of $\sqrt{x - 1} + \sqrt{5 - x}$.

Both square roots must be defined at the same time:

$$x - 1 \geq 0 \text{ gives } x \geq 1, \qquad 5 - x \geq 0 \text{ gives } x \leq 5$$

The com­mon val­ues are $x \in [1, 5]$. When a func­tion is a sum of sev­eral parts, its domain is the inter­sec­tion of the domains of the parts.

Exam­ple 8#

Find the domain of $\displaystyle \frac{1}{\sqrt{x^2 - 9}}$.

We need $x^2 - 9 > 0$, that is, $(x + 3)(x - 3) > 0$. Greater than zero, out­side the roots:

$$x \in (-\infty, -3) \cup (3, \infty)$$

Prac­tice#

Find the domain of each func­tion.

1. $\sqrt{x + 5}$
2. $\sqrt{6 - 2x}$
3. $\displaystyle \frac{1}{\sqrt{3x - 4}}$
4. $\sqrt{x^2 - 7x + 10}$
5. $\sqrt{9 - x^2}$
6. $\displaystyle \frac{1}{\sqrt{x^2 - 4x + 3}}$
7. $\sqrt{x - 2} + \sqrt{8 - x}$
8. $\sqrt{|x| - x}$
9. $\sqrt{x^2 + 1}$
10. $\displaystyle \frac{1}{\sqrt{x - |x|}}$
11. $\sqrt{|x| - 3}$

Answers#