Find Range of Square Root Functions

Find the range of square root functions; examples and matched problems with their answers at the bottom of the page.

Graphical Analysis of Range of Square Root Functions

The range of a function \( y = f(x) \) is the set of values \( y \) takes for all values of \( x \) within the domain of \( f \).
What is the range of \( f(x) = \sqrt{x} \) ?
The domain of \( f \) above is the set of all values of \( x \) in the interval [0 , \( +\infty \))
As \( x \) takes values from 0 to \( +\infty \), \( \sqrt{x} \) takes values from \( \sqrt {0} \) to \( \sqrt{\infty} \) (see blue graph below). Hence the range of \( \sqrt {x} \) is given by the interval: [0 , \( +\infty \))
The range of function of the form \( \sqrt {x + k} \) , (see red and green graphs below) is given by the interval:[0 , \( +\infty \)). Right and left shifts do not affect the range of function.
We can also write the range \( y = f(x) = \sqrt{x + k} \) of in the following inequality form
\( y \geq 0 \) or \( \sqrt{x + k} \geq 0 \)

Fig1. - Examples of Square Root Functions.

Examples with Solutions

Example 1

Find the range of function \( f \) defined by
\( f(x) = \sqrt {x - 1} \)

Solution to Example 1

We know, from the discussion above, that the range of function \( f(x) = \sqrt {x} \) is given by the interval [0 , \( +\infty \)).
The graph of the given function \( f(x) = \sqrt{x - 1} \) is the graph of \( \sqrt{x} \) shifted 1 unit to the right. A shift to the right does not affect the range. Hence the range of \( f(x) = \sqrt{x - 1} \) is also given by the interval: [ 0 , \( +\infty \) )

Matched Problem 1:

Find the range of function \( f \) defined by
\( f(x) = \sqrt {x + 5} \)

Example 2

Find the range of function \( f \) defined by
\( f(x) = - \sqrt {x + 2} \)

Solution to Example 2

We first start with the range of values of expression \( \sqrt {x + 2} \) which may be written in inequality form as follows
\( \sqrt {x + 2} \geq 0 \)
Multiply both sides of the inequality by -1 to obtain to obtain the inequality
\( - \sqrt {x + 2} \leq 0 \)
The range of the expression \( - \sqrt {x + 2} \) which is also the range of the given function is given by the interval ( \( -\infty \) , 0]

Matched Problem 2:

Find the range of function \( f \) defined by
\( f(x) = - \sqrt {x - 4} \)

Example 3

Find the range of function \( f \) defined by
\( f(x) = - 2 \sqrt {x + 3} + 5 \)

Solution to Example 3

The range of values of \( \sqrt{x + 3} \) may be written as an inequality
\( \sqrt{x + 3} \geq 0 \)
Multiply both sides by -2 ( and change the sign of inequality ) to obtain
\(- 2 \sqrt {x + 3} \leq 0 \)
Add 5 to both sides of the above inequality to obtain
\(- 2 \sqrt{x + 3} + 5 \leq 5\)
The range of values of the expression on the left side of the inequality, which is also the range of the given function, is given by the interval
( \( - \infty , 5 ]\)

Matched Problem 3:

Find the range of function \( f \) defined by
\( f(x) = - 5 \sqrt {x + 3} - 10 \)

Example 4

Find the range of function \( f \) defined by
\( f(x) = \sqrt{16 - x^2} \)

Solution to Example 4

We first need to find the domain of the given function defined as the values of \( x \) such that
\( 16 - x^2 \geq 0 \)
The solution set to the above inequality is the domain of \( f(x) \) and is given by the interval
[ -4 , 4 ]
The range of values of \( 16 - x^2 \) for \( x \) in the interval [ -4 , 4 ] (domain) is given by the interval [0 , 16] since the graph is a parabola with a maximum at the point (0 , 16).
The given function is the square root of \( 16 - x^2 \) and therefore has the range defined by the interval [ \( \sqrt{0} \) , \( \sqrt{16} \) ] = [ 0 , 4 ]. See graphs below for better understanding

Range of square root function — Fig2. - Range of Square Root Functions.

Matched Problem 4:

Find the range of function \( f \) defined by
\( f(x) = \sqrt{4 - x^2} \)

Example 5

Find the range of function \( f \) defined by
\( f(x) = \sqrt{x^2 - 25} \)

Solution to Example 5

The domain of the given function is the set of \( x \) values such that
\( x^2 - 25 \geq 0 \)
The solution set to the above inequality is the domain of \( f(x) \) and is given by the interval
( \( - \infty \) , -5] \( \cup \) [5 , \( + \infty \) )
For \( x \) in the interval ( \( - \infty \) , -5] \( \cup \) [5 , \( + \infty \) ) the range of the expression \( x^2 - 25 \) is given by the interval [ 0 , \( +\infty \) ). The given function is the square root of \( x^2 - 25 \). Hence the range of the given function is given by the interval
[ \( \sqrt{0} \) , \( \sqrt{+\infty} \) ) = [ 0 , \( +\infty \) ) . See graph below for better understanding.

Matched Problem 5:

Find the range of function \( f \) defined by
\( f(x) = \sqrt{x^2 - 1} \)

Example 6

Find the range of function \( f \) defined by
\( f(x) = \sqrt{x^2 - 4x + 8} \)

Solution to Example 6

The domain of the given function is the set of \( x \) values such that
\( x^2 - 4x + 8 \geq 0 \)
The discriminant of the quadratic expression \( x^2 - 4x + 8 \) is given by
\( (-4)^2 - 4 (1)(8) = -16 \)
Since the discriminant is negative, the quadratic expression is either positive or negative for all values of \( x \). A test with \( x = 0 \) reveals the expression \( x^2 - 4x + 8 \) is always positive and therefore the domain of the given function is the set of all real numbers.
We next find the range of the expression \( x^2 - 4x + 8 \) which may be written as
\( x^2 - 4x + 8 = (x - 2)^2 + 4 \)
The graphs of \( (x - 2)^2 + 4 \) is a parabola with a minimum at (2 , 4) (the vertex). Hence the range of \( x^2 - 4x + 8 \) is given by the interval [ 4 , \( +\infty \) ). The given function is the square root of \( x^2 - 4x + 8 \) and therefore has the range given by
[ \( \sqrt{4} \) , \( \sqrt{+\infty} \) ) = [ 2 , \( +\infty \) ). See graph below for better understanding.

Matched Problem 6:

Find the range of function \( f \) defined by
\( f(x) = \sqrt{x^2 + 2x + 10} \)

Answers to the Above Matched Problems

1) [0 , \( +\infty \))
2) ( \( -\infty \) , 0 ]
3) ( \( -\infty \) , -10 ]
4) [ 0 , 2 ]
5) [ 0 , \( +\infty \))
6) [ 3 , \( +\infty \))