So far, we have seen functions such as the square, f(x) = x2, and cube, f(x) = x3, of a number. These are known as polynomial functions where the x's are the base and the superscripts 2 and 3 are the exponents (or power) of the expression. In this article, we shall look at the inverse of these expressions, called the square root and cube root. These are known as radicals.

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The word radical stems from the Latin phrase 'radix' which is defined by the term 'root'. It is represented by the symbol '√', which is a stylized or deformed "r". In English, the word 'root' describes the source of a particular object.

For example, the terms 'abnormal' and 'paranormal' come from the root word 'normal'. The same concept applies to numbers as well. Take the square of 3 for instance: 32 = 9. Here, 3 is the root and 9 is the result from the square of that root.

A radical expression is an expression that includes a radical symbol √.

$\sqrt{2},\sqrt{\frac{1}{3}},\sqrt[3]{x},\sqrt[x]{y}$

Common radicals include the square root and cube root of a number, denoted by the expressions

$\sqrt{x}and\sqrt[3]{x}$.

Solving an expression with a square root is rather straightforward. Simply take the square root of the radicand. The square root of a number is the number equal to the radicand when squared or multiplied by itself.

$\sqrt{81}=9\mathrm{sin}ce9×9=81$

Similarly, the cube root of a number is the number equal to the radicand when cubed or multiplied by itself three times.

$\sqrt[3]{125}=5\mathrm{sin}ce5×5×5=125$

Common radical functions include the square root function and cube root function defined by

$f\left(x\right)=\sqrt{x}$ and $f\left(x\right)=\sqrt[3]{x}$

respectively. Other forms of rational functions include

$f\left(x\right)=\sqrt{2x-1},\phantom{\rule{0ex}{0ex}}g\left(x\right)=\sqrt[4]{7{x}^{2}+3,}\phantom{\rule{0ex}{0ex}}h\left(x\right)=\sqrt[5]{{\left(2-{x}^{3}\right)}^{2}},etc$

Evaluating radical functions is similar to solving regular functions. Simply substitute the given x value into our function to find the value of f(x). Below are several worked examples demonstrating this.

$f\left(x\right)=\sqrt{3-4x}$

Evaluate

$f\left(-1\right),f\left(0\right),f\left(\frac{1}{2}\right)$

Solutions

$f\left(-1\right)=\sqrt{3-4\left(-1\right)}=\sqrt{3+4}=\sqrt{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}f\left(0\right)=\sqrt{3-4\left(0\right)}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}f\left(\frac{1}{2}\right)=\sqrt{3-4\left(\frac{1}{2}\right)}=\sqrt{3-2}=\sqrt{1}=1$

$g\left(x\right)=\sqrt[3]{2{x}^{2}+5}$

Evaluate

$g\left(-2\right),g\left(\frac{5}{2}\right),g\left(7\right)$

Solutions

$g\left(-2\right)=\sqrt[3]{2{\left(-2\right)}^{2}+5}=\sqrt[3]{8+5}=\sqrt[3]{13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}g\left(\frac{5}{2}\right)=\sqrt[3]{2{\left(\frac{5}{2}\right)}^{2}+5}=\sqrt[3]{\frac{25}{2}+5}=\sqrt[3]{\frac{35}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}g\left(7\right)=\sqrt[3]{2{\left(7\right)}^{2}+5}=\sqrt[3]{98+5}=\sqrt[3]{103}$

## Graphs and Transformations of Radical Functions

As mentioned before, the radical functions y = √x and y = 3√x are the inverses of the polynomial functions y = x2 and y = x3, respectively. In this section, we shall explore these functions further by comparing the shape of their graphs and their related transformations. Additionally, we shall study the values of their domain and range.

### Finding the Domain and Range of a Radical Function

Recall the definitions of a domain and range of a function.

The domain of a function f is the set of all possible values x such that f is defined.

The range of a function f is the set of all possible values f can take, where x is any number from the domain of f.

To identify the domain and range of a radical function, let us recall the characteristics of the square root and cube root function below.

### The Square Root Function

The square root function is given by

$f\left(x\right)=\sqrt{x}$

The graph of the function is illustrated below.

The square root function, Aishah Amri - StudySmarter Originals

On the same graph, let us graph the inverse of this function, y = x2. This is represented by the blue curve.

The square root and square function, Aishah Amri - StudySmarter Originals

Let us compare the two curves. Notice how the two graphs are reflected about the line x = y which indicates that these two functions are inverses of each other.

Now observe the graph of y = √x. Given the square root function, for a real number y to satisfy y = f(x), we must have x = y2. The square of any real number is non-negative, and so x must also be non-negative. In other words, we cannot take the square root of a negative real number. Thus, the domain of the square root function is x ≥ 0. In interval notation, this is denoted by [0, ∞).

Since the domain does not include negative real numbers, it follows that the range also considers positive real numbers only. The square root of positive real numbers defined by the domain is indeed positive as well. Again, the range is expressed by the interval notation [0, ∞). That's why the graph shows a half parabola! From this plot, we can draw the following conclusion.

### The Domain and Range of a Square Root Function

The domain and range of the square root function

$f\left(x\right)=\sqrt{x}$

is [0, ∞). In essence, the domain of the composite square root function

$y=\sqrt{g\left(x\right)}$

can be identified by finding the values of x satisfying

$g\left(x\right)\ge 0$ .

If the range of the g is the set of all real numbers, then the range of$\sqrt{g\left(x\right)}$is [0, ∞). This idea is demonstrated by the examples below.

Given the function below, find the domain and range of the function.

$f\left(x\right)=\sqrt{2x-5}$

Solution

By the statement above, given that:

$g\left(x\right)=2x-5$

The domain of this function is the set of x-values such that:

$g\left(x\right)\ge 0⇒2x-5\ge 0$

Rearranging this inequality, we obtain the domain of the function as:

$2x\ge 5⇒x\ge \frac{5}{2}$

Since the range of g is the set of all real numbers, it follows that the range of f is [0, ∞). The graph of this function is shown below.

Example 1, Aishah Amri - StudySmarter Originals

Given the function below, find the domain and range of the function.

$f\left(x\right)=\sqrt{3-4x}$

Solution

As before, setting g as:

$g\left(x\right)=3-4x$

The domain of this function is the set of x-values such that:

$g\left(x\right)\ge 0⇒3-4x\ge 0$

Rearranging this inequality, we obtain the domain of the function as

$3\ge 4x⇒x\le \frac{3}{4}$

Since the range of g is the set of all real numbers, it follows that the range of f is [0, ∞). The plot of this function is displayed below.

Example 2, Aishah Amri - StudySmarter Originals

### Transformation of the Square Root Function

In this section, we shall look at the square root function of the form

$y=a\sqrt{x-h}+k$

with respect to its standard form $y=\sqrt{x}.$ Here, we shall observe how varying the values of a, h, and k will affect the shape of the curve. The table below describes this in detail. In each plot below, the graph of $y=\sqrt{x}$ is drawn in red.

 Change of Value in $y=a\sqrt{x-h}+k$ Description Graphical Representation Varying a changes the function in the y-direction (the coefficient a affects the steepness of the graph) If a is large, the graph becomes steeper (orange line)If a is small, the graph becomes flatter (blue line)If a is negative, the graph becomes inverted (green line) Change in a, Aishah Amri - StudySmarter Originals Varying h changes the function along the x-axis by h units If h is negative, the graph shifts h units to the left of the x-axis (green line)If h is positive, the graph shifts h units to the right of the x-axis (blue line) Change in h, Aishah Amri - StudySmarter Originals Varying k moves the function up or down the y-axis by k units If k is negative, the graph moves down k units in the y-axis (green line)If k is positive, the graph moves up k units in the y-axis (blue line) Change in k, Aishah Amri - StudySmarter Originals

Graph the function of $y=2\sqrt{x-3}+1$

Solution

Here, a = 2. This implies that although the graph is of the same shape as the standard square root graph (with a = 1), the curve happens to be steeper since 2 is greater than 1. Furthermore, h = 3 and k = 1 which infers that the graph must shift 3 units to the right and 1 unit up. The graph is shown below.

Example 3, Aishah Amri - StudySmarter Originals

### The Cube Root Function

The cube root function is given by

$f\left(x\right)=\sqrt[3]{x}$

Below is an illustration of this function.

The cube root function, Aishah Amri - StudySmarter Originals

On the same graph, let us graph the inverse of this function, y = x3. This is represented by the blue curve.

The cube root and cube function, Aishah Amri - StudySmarter Originals

We shall now compare the two curves. Again, the two graphs are reflected about the line x = y. This indicates that this pair of functions are inverses of each other.

Let us return to the plot of y = 3√x. The graph shows that the domain and range are given by the set of all real numbers. Observe that the function tends to positive infinity as the curve moves to the right and negative infinity as the curve moves to the left. Unlike the square root function, the cube root function does not have any restrictions on the domain and range. In interval notation, this is denoted by (-∞, ∞).

### The Domain and Range of a Cube Root Function

The domain and range of the cube root function

$f\left(x\right)=\sqrt[3]{x}$

is given by the set of all real numbers and is denoted by (-∞, ∞). We may also represent this set by the notation$\mathrm{ℝ}$. This means that the domain and range of the composite cube root function

$y=\sqrt[3]{g\left(x\right)}$

is also denoted by interval notation (-∞, ∞). Let us observe some worked examples below.

Given the function below, find the domain and range of the function.

$f\left(x\right)=\sqrt[3]{7x-2}$

Solution

As mentioned before, the cube root function does not impose any restrictions on the domain and range of any composite cube root function. Thus, the domain and range of f is given by (-∞, ∞). The graph of this function is shown below.

Example 4, Aishah Amri - StudySmarter Originals

Given the function below, find the domain and range of the function.

$f\left(x\right)=\sqrt[3]{9-3x}$

Solution

Similar to the result above, the cube root function does not impose any restrictions on the domain and range of any composite cube root function. Thus, the domain and range of f is given by (-∞, ∞). The plot of this function is displayed below.

Example 5, Aishah Amri - StudySmarter Originals

### Transformation of the Cube Root Function

We shall move on to investigating the cube root function of the form

$y=a\sqrt[3]{x-h}+k$

with respect to its standard form $y=\sqrt[3]{x}.$ Similar to the square root case, we seek to study how changing the values of a, h, and k will influence the shape of the curve. The table below shows this in detail. In each plot below, the graph of $y=\sqrt[3]{x}$ is drawn in red.

 Change in Variable $y=a\sqrt[3]{x-h}+k$ Description Graphical Representation Varying a changes the function in the y-direction (the coefficient a affects the steepness of the graph) If a is large, the graph becomes steeper (orange line)If a is small, the graph becomes flatter (blue line)If a is negative, the graph becomes inverted (green line) Change in a, Aishah Amri - StudySmarter Originals Varying h changes the function along the x-axis by h units If h is negative, the graph shifts h units to the left of the x-axis (green line)If h is positive, the graph shifts h units to the right of the x-axis (blue line) Change in h, Aishah Amri - StudySmarter Originals Varying k moves the function up or down the y-axis by k units If k is negative, the graph moves down k units in the y-axis (green line)If k is positive, the graph moves up k units in the y-axis (blue line) Change in k, Aishah Amri - StudySmarter Originals

Graph the function of $y=\frac{1}{4}\sqrt[3]{x+2}-7$

Solution

Here, $a=\frac{1}{4}$. This implies that although the graph is of the same shape as the standard cube root graph (with a = 1), the curve happens to be flatter since $\frac{1}{4}$ is greater than 1. Furthermore, h = -2 and k = -7 which infers that the graph must shift 2 units to the left and 7 units down. The graph is shown below.

Example 6, Aishah Amri - StudySmarter Originals

## Solving Radicals Equations and Inequalities

Solving radical equations and inequalities are similar to solving regular equations. To solve such an expression, simply raise both sides of the equation to the power equal to the index of the radical. This eliminates the radical. For instance, to get rid of a square root, you must square both sides of the expression. Similarly, to undo an nth root, you must raise the expression to the nth power.

It is very important that you verify your solution. Often, you will obtain a number that does not satisfy the original equation or inequality, meaning that a solution does not exist. Below are some examples that demonstrate this.

Solve the equation below.

$5=\sqrt{x-2}-1$

Solution

Isolating the radical and squaring both sides, we obtain:

$5=\sqrt{x-2}-1\phantom{\rule{0ex}{0ex}}⇒\sqrt{x-2}=6\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{x-2}\right)}^{2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒x-2=36\phantom{\rule{0ex}{0ex}}⇒x=38$

Check that this result is true:

$\sqrt{x-2}-1=\sqrt{\left(38\right)-2}-1=\sqrt{36}-1=6-1=5$

Thus, x = 38 satisfies the given equation.

Solve the equation below.

$\sqrt{x+15}=5+\sqrt{x}$

Solution

Squaring both sides and isolating the radical, we obtain:

$\sqrt{x+15}=5+\sqrt{x}\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{x+15}\right)}^{2}={\left(5+\sqrt{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒x+15=25+10\sqrt{x}+x\phantom{\rule{0ex}{0ex}}⇒10\sqrt{x}=-10\phantom{\rule{0ex}{0ex}}⇒\sqrt{x}=-1$

Squaring both sides again, we obtain:

$\sqrt{x}=-1\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{x}\right)}^{2}={\left(-1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒x=1$

Check that this result is true:

The left-hand side is given by:

$\sqrt{x+15}=\sqrt{\left(1\right)+15}=\sqrt{16}=4$

The right-hand side is given by:

$5+\sqrt{x}=5+\sqrt{1}=5+1=6$

Thus, 4 ≠ 6. Since the left-hand side is not equal to the right-hand side, x = 1 does not satisfy the equation. In other words, no solution exists.

Solve the inequality below.

$\sqrt{2x+2}+1\ge 5$

Solution

Isolating the radical and squaring both sides, we obtain:

$\sqrt{2x+2}+1\ge 5\phantom{\rule{0ex}{0ex}}⇒\sqrt{2x+2}\ge 4\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{2x+2}\right)}^{2}\ge {4}^{2}\phantom{\rule{0ex}{0ex}}⇒2x+2\ge 16\phantom{\rule{0ex}{0ex}}⇒2x\ge 14\phantom{\rule{0ex}{0ex}}⇒x\ge 7$

Check that this result is true:

$\sqrt{2x+2}+1=\sqrt{2\left(7\right)+2}+1=\sqrt{16}+1=4+1=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{2x+2}+1=\sqrt{2\left(8\right)+2}+1=\sqrt{18}+1=3\sqrt{2}+1\ge 5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{2x+2}+1=\sqrt{2\left(9\right)+2}+1=\sqrt{20}+1=2\sqrt{5}+1\ge 5$

Thus, x ≥ 7 satisfies the given inequality.

Solve the inequality below.

$\sqrt{4x-4}-2<4$

Solution

Isolating the radical and squaring both sides, we obtain:

$\sqrt{4x-4}-2<4\phantom{\rule{0ex}{0ex}}⇒\sqrt{4x-4}<6\phantom{\rule{0ex}{0ex}}⇒{\left(\sqrt{4x-4}\right)}^{2}<{6}^{2}\phantom{\rule{0ex}{0ex}}⇒4x-4<36\phantom{\rule{0ex}{0ex}}⇒4x<40\phantom{\rule{0ex}{0ex}}⇒x<10$

Recall that we cannot take the square root of a negative real number. Thus, the term inside the radical must satisfy

$4x-4\ge 0$

Solving this, we obtain:

$x\ge 1$

Thus, 1 ≤ x < 10. Check that this result is true:

$\sqrt{4x-4}-2=\sqrt{4\left(1\right)-4}-2=\sqrt{0}-2=-2<4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{4x-4}-2=\sqrt{4\left(5\right)-4}-2=\sqrt{16}-2=4-2=2<4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{4x-4}-2=\sqrt{4\left(10\right)-4}-2=\sqrt{36}-2=6-2=4\nless 4$

Thus, the correct range of x that satisfies the given inequality is 1 ≤ x < 10.

## Radical Functions - Key takeaways

• A radical expression is an expression that includes a radical symbol (√).
• A radical function is a function that contains a radical expression.
• The domain and range of a square root function is [0, ∞).
• The domain of the composite square root function$\sqrt{g\left(x\right)}$are the values of x satisfying $g\left(x\right)\ge 0$.
• The domain and range of a cube root function is (-∞, ∞).
• The domain of the composite cube root function $\sqrt[3]{g\left(x\right)}$ is $\mathrm{ℝ}$.
• If the range if the function g is the set of all real numbers, then the range of the composite functions $\sqrt{g\left(x\right)}$ and $\sqrt[3]{g\left(x\right)}$ is [0, ∞) and (-∞, ∞) respectively.
• To solve radical equations and inequalities:
1. Eliminate the radical by raising both sides of the expression to the nth power.
2. Solve the equation or inequality.
3. Check that the answer satisfies the expression.

#### Flashcards in Radical Functions 15

###### Learn with 15 Radical Functions flashcards in the free StudySmarter app

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A radical function is a function that includes a square root

What is the formula for a radical function

The formula for a radical function is to solve the nth root

How do you solve and graph radical functions

Isolate the radical to one side of the equation, raise both sides of the equation to the power of the index and solve the equation

What is the first step in solving a radical equation

Isolate the radical to one side of the equation and raise both sides of the equation to the power of the index

What is an example of radical functions

An example of a radical function is the square root function f(x) = x

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