MAT 142: Radical Functions and Equations

Reminders and Warm-Up

Last class we studied rational functions. Before moving on, let’s practice solving a rational equation — and encounter a trap that will reappear today.

Problem 1: Solve \(\displaystyle{\frac{3}{x} + 2 = \frac{5}{x}}\).

Hint. To start, multiply both sides by \(x\) (the simplest common denominator), then solve.

Problem 2: Solve \(\displaystyle{\frac{6}{x^2 - 9} + \frac{1}{x + 3} = \frac{2}{x - 3}}\).

Hint. Start similarly and pay close attention to your solutions.

Extraneous Solutions

An extraneous solution is an algebraically valid result that fails when substituted back into the original equation — often because it violates a domain restriction. Always check solutions when multiplying by an expression containing a variable.

This same trap appears today with radical equations, where squaring both sides can introduce extraneous solutions.

Objectives

Today we study radical functions — functions built from roots of expressions.

After today’s class meeting, you should be able to:

Identify the domain of a radical function based on the index of the root.
Solve polynomial inequalities using sign testing.
Find the roots and \(y\)-intercept of a radical function.
Sketch the graph of a radical function.

What Is a Radical Function?

Definition (Radical Function): A function of the form

\[f\left(x\right) = \sqrt[n~~~]{g\left(x\right)} = \left(g\left(x\right)\right)^{1/n}\]

where \(n\) is a positive integer, is called a radical (or root) function.

Domain depends on the index \(n\):

If \(n\) is odd: the domain is all real numbers (odd roots of negatives are defined).
If \(n\) is even: the domain requires \(g\left(x\right) \geq 0\) (even roots of negatives result in complex numbers).

Roots: \(f\left(x\right) = 0\) only when \(g\left(x\right) = 0\).

\(y\)-intercept: Evaluate \(f\left(0\right)\), provided \(0\) is in the domain.

On Notation

While the root notation (\(\sqrt{~~~}\)) is common, it is almost always more mathematically convenient to use the exponent notation.

For example, instead of witing \(f\left(x\right) = \sqrt{x}\), write \(f\left(x\right) = x^{1/2}\)

Domain of Radical Functions

Example 1: Find the domain of the radical function \(\displaystyle{f\left(x\right) = 3\sqrt{\left(x - 9\right)} + 17}\)

Solution.

Since the order of the root is even, we need the argument (the expression inside of the square root) to be non-negative.

\[\begin{align} x - 9\geq 0 \end{align}\]

Domain of Radical Functions

Example 1: Find the domain of the radical function \(\displaystyle{f\left(x\right) = 3\sqrt{\left(x - 9\right)} + 17}\)

Solution.

Since the order of the root is even, we need the argument (the expression inside of the square root) to be non-negative.

\[\begin{align} x - 9\geq 0\\ \implies x \geq 9 \end{align}\]

The domain of \(f\left(x\right)\) is \(x\geq 9\), or equivalently \(\left[9, \infty\right)\)

Example 2: Find the domain of the radical function \(\displaystyle{g\left(x\right) = 5\left(\frac{x + 8}{7}\right)^{1/7} - 4}\).

Solution.

Since the order of the root is odd (it is a seventh root), the domain consists of all real numbers, or equivalently \(\left(-\infty, \infty\right)\).

Domain of Radical Functions Practice

Before encountering a comprehensive example, let’s build intuition for domains. Find the domain of each function:

Try It! 1: Find the domain of the function \(\displaystyle{f\left(x\right) = \sqrt{\left(\frac{3x + 4}{8}\right)}}\).

Try It! 2: Find the domain of the function \(\displaystyle{g\left(x\right) = 3\sqrt[5~]{\left(x^2 - 9x + 20\right)}}\).

Try It! 3: Find the domain of the function \(\displaystyle{h\left(x\right) = -2\left(x^2 - 9x + 20\right)^{1/4} + 10}\).

Try It! 4: Find the domain of the function \(\displaystyle{j\left(x\right) = \frac{5}{\sqrt{\left(8 - 2x\right)}}}\)

There are several interesting things to consider across the examples above.

The second and third functions have the same expression inside the radical. Why do the domains differ?
Why must the inequality describing the domain in of the fourth function be strict (that is, it doesn’t include the finite boundary endpoint)?
What is an appropriate strategy for solving the inequality describing the domain of the third function?

Solving Polynomial Inequalities

Finding domains of even-degree radical functions requires solving inequalities like \(g\left(x\right) \geq 0\). In all but the simplest cases (like Example 1 and Try It! 1), we use sign testing.

Strategy (Sign Testing): To find the domain of a function that includes \(\displaystyle{\left(g\left(x\right)\right)^{\frac{1}{2n}}}\), follow the steps below.

Find the roots of \(g\left(x\right)\) — these are the potential boundary locations.
The roots divide the number line into intervals.
Test one point in each interval to determine the sign of \(g\left(x\right)\) there.
The domain for this part of the function consists of intervals where the sign is positive (or zero, if it is permissible for \(\displaystyle{\left(g\left(x\right)\right)^{\frac{1}{2n}}}\) to evaluate to \(0\)).

The function can only change sign at its roots, so one test point per interval is sufficient.

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\)

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0 \end{align}\]

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0 \end{align}\]

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Can either evaluate the sign (positive/negative) from \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 3: Find the domain of the radical function \(\displaystyle{f\left(x\right) = \left(x^2 -5x - 14\right)^{1/8}}\)

Solution.

Since the root is even order (8th root), we need to consider domain restrictions.

The function \(f\left(x\right)\) will only be defined when \(x^2 - 5x - 14 \geq 0\).

We’ll start by finding the boundary values where \(x^2 - 5x - 14 = 0\).

\[\begin{align} x^2 - 5x - 14 &= 0\\ \implies \left(x - 7\right)\left(x + 2\right) &= 0\\ \implies x = 7 &\text{ or } x = -2 \end{align}\]

Conduct Sign Analysis:

Either evaluate the sign (positive/negative) using \(x^2 - 5x - 14\) or \(\left(x - 7\right)\left(x + 2\right)\).

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[7, \infty\right)~}\)

The analysis above results in three sub-intervals where the expression under the root is not zero: \(\left(-\infty, -2\right)\), \(\left(-2, 7\right)\), and \(\left(7, \infty\right)\).

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

More Domains of Radical Functions

Example 4: Find the domain of the radical function \(\displaystyle{g\left(x\right) = \left(\left(x + 5\right)\left(x + 1\right)\left(x - 2\right)^2\left(x - 6\right)\right)^{1/2}}\)

Solution.

Again, we need the expression under the root to be non-negative.

Notice that the boundary values (the \(x\)-values resulting in \(0\) under the radical) are \(x = -5\), \(-1\), \(2\), and \(6\)

We’ll draw a number line and conduct the sign analysis.

All of the boundary values and the segments corresponding to positive values under the radical are included in the domain.

We’ll use \(\{2\}\) to indicate the single value \(2\) when we write the domain.

Domain: \(\boxed{~\left[-5, -1\right]\cup\left\{2\right\}\cup\left[6, \infty\right)~}\)

Note. As seen here, intervals won’t necessarily alternate signs. Test them all.

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = \left(-10\right)^2 - 11\left(-10\right) - 26}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = \left(100 + 110 - 26\right)}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = \left(0\right)^2 - 11\left(0\right) - 26}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = - 26}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = - 26}\)

\(\displaystyle{g\left(15\right) = \left(15\right)^2 - 11\left(15\right) - 26}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = - 26}\)

\(\displaystyle{g\left(15\right) = 225 - 165 - 26}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = - 26}\)

\(\displaystyle{g\left(15\right) = 34}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Since the index is \(2\) (even), we need \(x^2 - 11x - 26 \geq 0\).

Step 1: Find roots of \(x^2 - 11x - 26 = 0\).

\[\begin{align} x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = 13 &\text{ or } x = -2 \end{align}\]

Step 2: These roots create three intervals

\(\left(-\infty, -2\right)\), \(\left(-2, 13\right)\), and \(\left(13, \infty\right)\)

Step 3: Test one point in each interval.

\(\displaystyle{g\left(-10\right) = 184}\)

\(\displaystyle{g\left(0\right) = - 26}\)

\(\displaystyle{g\left(15\right) = 34}\)

Domain: \(\boxed{~\left(-\infty, -2\right] \cup \left[13, \infty\right)~}\)

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: We’ll solve \(f\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(x^2 - 11x - 26\right)} &= 0 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: We’ll solve \(f\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(x^2 - 11x - 26\right)} &= 0\\ \implies x^2 - 11x - 26 &= 0 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: We’ll solve \(f\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(x^2 - 11x - 26\right)} &= 0\\ \implies x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: We’ll solve \(f\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(x^2 - 11x - 26\right)} &= 0\\ \implies x^2 - 11x - 26 &= 0\\ \implies \left(x - 13\right)\left(x + 2\right) &= 0\\ \implies x = -2 &\text{ or } x = 13 \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: The roots of \(f\left(x\right)\) are at \(\boxed{~\left(-2, 0\right)~}\) and \(\boxed{~\left(13, 0\right)~}\)

\(y\)-intercept: Note that \(x = 0\) is not in the domain of the function. This means that there is no \(y\)-intercept.

End behavior: We’ll analyse \(\displaystyle{\lim_{x\to -\infty}{f\left(x\right)}}\) and \(\displaystyle{\lim_{x\to \infty}{f\left(x\right)}}\)

\[\begin{align} \lim_{x\to -\infty}{f\left(x\right)} = \lim_{x\to -\infty}{\sqrt{\left(x^2 - 11x - 26\right)}} \end{align}\]

The quadratic under the root goes off to infinity.

Because of this, the root goes off to infinity as well.

\[\begin{align} \lim_{x\to -\infty}{f\left(x\right)} &= \infty \end{align}\]

\[\begin{align} \lim_{x\to \infty}{f\left(x\right)} = \lim_{x\to \infty}{\sqrt{\left(x^2 - 11x - 26\right)}} \end{align}\]

The quadratic under the root goes off to infinity.

Because of this, the root goes off to infinity as well.

\[\begin{align} \lim_{x\to \infty}{f\left(x\right)} &= \infty \end{align}\]

Analysis of Radical Functions I

Example 5: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 11x - 26\right)}}\). Use your results to sketch the graph of \(f\left(x\right)\)

Solution.

Domain: \(\boxed{~\left(-\infty, -2\right]\cup \left[13, \infty\right)~}\)

Roots: The roots of \(f\left(x\right)\) are at \(\boxed{~\left(-2, 0\right)~}\) and \(\boxed{~\left(13, 0\right)~}\)

\(y\)-intercept: Note that \(x = 0\) is not in the domain of the function. This means that there is no \(y\)-intercept.

End behavior: We’ve found \(\displaystyle{\lim_{x\to -\infty}{f\left(x\right)} = \infty}\) and \(\displaystyle{\lim_{x\to \infty}{f\left(x\right)} = \infty}\)

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0\\ \implies x^2 &= 16 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0\\ \implies x^2 &= 16\\ \implies x &= \pm 4 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0\\ \implies x^2 &= 16\\ \implies x &= \pm 4 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0\\ \implies x^2 &= 16\\ \implies x &= \pm 4 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: We’ll start by finding the boundary values.

\[\begin{align} 16 - x^2 &= 0\\ \implies x^2 &= 16\\ \implies x &= \pm 4 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: We’ll solve \(g\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(16 - x^2\right)} &= 0 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: We’ll solve \(g\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(16 - x^2\right)} &= 0\\ \implies 16 - x^2 &= 0 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: We’ll solve \(g\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(16 - x^2\right)} &= 0\\ \implies 16 - x^2 &= 0\\ \implies x^2 = 16 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: We’ll solve \(g\left(x\right) = 0\)

\[\begin{align} \sqrt{\left(16 - x^2\right)} &= 0\\ \implies 16 - x^2 &= 0\\ \implies x^2 = 16\\ \implies x = \pm 4 \end{align}\]

There’s no surprise here. The roots are the boundary values we found earlier.

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: The roots of \(g\left(x\right)\) are at \(\left(-4, 0 \right)\) and \(\left(4, 0\right)\).

\(y\)-intercept: We’ll evaluate \(g\left(0\right)\).

\[\begin{align} g\left(0\right) = \sqrt{\left(16 - \left(0\right)^2\right)} \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: The roots of \(g\left(x\right)\) are at \(\left(-4, 0 \right)\) and \(\left(4, 0\right)\).

\(y\)-intercept: We’ll evaluate \(g\left(0\right)\).

\[\begin{align} g\left(0\right) &= \sqrt{\left(16 - \left(0\right)^2\right)}\\ &= \sqrt{16} \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: The roots of \(g\left(x\right)\) are at \(\left(-4, 0 \right)\) and \(\left(4, 0\right)\).

\(y\)-intercept: We’ll evaluate \(g\left(0\right)\).

\[\begin{align} g\left(0\right) &= \sqrt{\left(16 - \left(0\right)^2\right)}\\ &= \sqrt{16}\\ &= 4 \end{align}\]

Analysis of Radical Functions II

Example 6: Find the domain, roots, \(y\)-intercept, and end behavior of \(\displaystyle{g\left(x\right) = \sqrt{\left(16 - x^2\right)}}\). Use your results to sketch the graph of \(g\left(x\right)\).

Solution.

Domain: \(\left[-4, 4\right]\)

Roots: The roots of \(g\left(x\right)\) are at \(\left(-4, 0 \right)\) and \(\left(4, 0\right)\).

\(y\)-intercept: The \(y\)-intercept of the function is at \(\left(0, 4\right)\).

End Behavior: Not applicable since the domain does not extend toward positive or negative infinity.

Radical Function Practice

For each radical function below, find: (i) the domain, (ii) any roots, and (iii) the \(y\)-intercept (if one exists).

Try It! 1: \(\displaystyle{f\left(x\right) = \sqrt{\left(4 - 2x\right)}}\)

Try It! 2: \(\displaystyle{g\left(x\right) = \sqrt{\left(3x^2 - 4x - 15\right)}}\)

Try It! 3: \(\displaystyle{h\left(x\right) = \sqrt{\left(x^2 - 9\right)}}\)

Try It! 4: \(\displaystyle{j\left(x\right) = \sqrt{\left(x^2 + 9\right)}}\)

Try It! 5: \(\displaystyle{k\left(x\right) = \sqrt{\left(x + 3\right)^2}}\)

Challenge for Try It! 5: Before computing, predict what the graph will look like. Then simplify \(\sqrt{\left(x+3\right)^2}\) algebraically. Does the result match your prediction?

Radical Functions and Inverse Functions

Radical functions arise naturally as inverses of power functions.

For example, \(f\left(x\right) = x^2\) is not invertible over all of \(\mathbb{R}\) — it fails the horizontal line test. But if we restrict its domain to \(x \geq 0\), then \(f\left(x\right) = x^2\) is invertible, and its inverse is \(f^{-1}\left(x\right) = \sqrt{x}\).

This is one justification for why \(\sqrt{x}\) is only defined for \(x \geq 0\). It’s the inverse of a function whose restricted domain was \(x \geq 0\).

More generally, if \(n\) is even, then \(f\left(x\right) = x^n\) restricted to \(x \geq 0\) has inverse \(f^{-1}\left(x\right) = x^{1/n} = \sqrt[n]{x}\).

For odd \(n\), no domain restriction is needed because \(f\left(x\right) = x^n\) already passes the horizontal line test on all of \(\mathbb{R}\).

Exit Ticket Task

Navigate to our MAT142 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 15. Radical Functions and Equations

Task: Consider \(\displaystyle{f\left(x\right) = \sqrt{\left(x^2 - 4x - 12\right)}}\).

\(\left(a\right)\) Find the domain of \(f\left(x\right)\) using sign testing.

\(\left(b\right)\) Find any roots and the \(y\)-intercept (if one exists).

\(\left(c\right)\) Sketch a rough graph.

Summary and Next Time…

Ideas From Today

A radical function \(f\left(x\right) = \left(g\left(x\right)\right)^{1/n}\) has domain that depends on whether \(n\) is even or odd.
- Odd index: all real numbers.
- Even index: solve \(g\left(x\right) \geq 0\) using sign testing.
Sign testing: find roots of \(g\), test one point per interval, keep intervals where the sign is non-negative.
Roots of \(f\) occur where \(g\left(x\right) = 0\).
Radical functions arise as inverses of power functions with domain restrictions.
Extraneous solutions can appear whenever you multiply both sides by a variable expression (rational equations) or raise both sides to a power (radical equations) — always check solutions.

Looking Ahead

With algebraic functions complete, we move into exponential and logarithmic functions.
These are genuinely new classes of function. For example, exponential functions have the independent variable in the exponent.
The inverse relationship between exponentials and logarithms connects directly back to our Day 9 discussion on inverse functions.
- Take a look back at that discussion for a refresher.

Next Time:
Basics of Exponentials and Logarithms

Homework:
Complete Homework 10 on MyOpenMath