In Sections ConstantandLinearFunctions, AbsoluteValueFunctions and QuadraticFunctions, we studied constant, linear, absolute value, and quadratic functions. Constant, linear and quadratic functions were specific examples of polynomial functions, which we studied in generality in Chapter PolynomialFunctions. Chapter PolynomialFunctions culminated with the Real Factorization Theorem, Theorem realfactorization, which says that all polynomial functions with real coefficients can be thought of as products of linear and quadratic functions. Our next step was to enlarge our field of study to rational functions in Chapter RationalFunctions. Being quotients of polynomials, we can ultimately view this family of functions as being built up of linear and quadratic functions as well. So in some sense, Sections ConstantandLinearFunctions, AbsoluteValueFunctions and QuadraticFunctions along with Chapters PolynomialFunctions and RationalFunctions can be thought of as an exhaustive study of linear and quadratic functions. We now turn our attention to functions involving radicals which cannot be written in terms of linear functions. For a more detailed review of the basics of roots and radicals, we refer the reader to Sections AppRealNumberArithmetic and AppRadEqus.

1 Root Functions

As with polynomial functions and rational functions, we begin our study of functions involving radical with a special family of functions: the (principal) root functions.

The domain restriction for even indexed roots means that, once again, we are restricting our attention to real numbers. We graph a few members of the root function family below using Desmos. As with the monomial, and, more generally, the Laurent monomial functions, the behavior of the root functions depends primarily on whether the root is even or odd.

As we select the graphs for indices \(n = 2\), \(n=4\), and \(n=6\), we observe that in addition to each function having the common domain \([0, \infty )\), the graphs all share the points \((0,0)\) and \((1,1)\). As the index \(n\) increases, the functions become ‘steeper’ near the \(y\)-axis and ‘flatter’ as \(x \rightarrow \infty \). These characteristics are common to all functions of the form \(f(x) = \sqrt [n]{x}\).

While the graphs appear to ‘flatten out’ as \(x \rightarrow \infty \), it turns out that \(f(x) \rightarrow \infty \) as \(x \rightarrow \infty \). First note that if \(c \geq 0\) is a real number, then \(f(c^n) = \sqrt [n]{c^n} = c\) so \(c\) is in the range of \(f\). This shows the range of \(f\) is \([0, \infty )\). Next note that \(f\) is increasing: that is, if \(a<b\), then \(f(a) = \sqrt [n]{a} < \sqrt [n]{b} = f(b)\). (This property is useful in solving certain types of polynomial inequalities. ) Hence, we get that \(\ds {\lim _{x \rightarrow \infty } f(x) = \infty }\).

The functions \(f(x) = \sqrt [n]{x}\) for odd natural numbers \(n \geq 3\) also follow a predictable trend - steepening near \(x = 0\) and flattening as \(x \rightarrow -\infty \) and \(x \rightarrow -\infty \). The range for these functions is \((-\infty , \infty )\) since if \(c\) is any real number, \(f(c^n) = \sqrt [n]{c^n} = c\), so \(c\) is in the range of \(f\). Like the even indexed roots, the odd indexed roots are also increasing. This gives us that \(\ds {\lim _{x \rightarrow \infty } f(x) = \infty }\) and \(\ds {\lim _{x \rightarrow -\infty } f(x) = -\infty }\).

Moreover, these graphs appear to be symmetric about the origin. Sure enough, when \(n\) is odd, \(f(-x) = \sqrt [n]{-x} = -\sqrt [n]{x} = -f(x)\) so \(f\) is an odd function.

At this point, you’re probably expecting a theorem like Theorems ??, ??, ??, ?? - that is, a theorem which tells us how to obtain the graph of \(F(x) = a \sqrt [n]{x-h}+k\) from the graph of \(f(x) = \sqrt [n]{x}\) - and you would not be wrong. Here, however, we need to add an extra parameter ‘\(b\)’ to the recipe and discuss functions of the form \(F(x) = a \sqrt [n]{bx-h}+k\). The reason is that, with all of the previous function families, we were always able to factor out the coefficient of \(x\). We list some examples of this below, and invite the reader to revisit other examples in the text:

  • \(F(x) = |6-2x| = |-2x+6| = |-2(x+3)| = |-2||x+3| = 2 |x+3|\).
  • \(F(x) = (2x-1)^2 + 1 = \left [2 \left (x - \frac {1}{2}\right )\right ]^2+1 = (2)^2 \left (x - \frac {1}{2}\right )^2 + 1 = 4\left (x - \frac {1}{2}\right )^2 + 1\)
  • \(F(x) = \frac {2}{(1-x)^3}- 5 = \frac {2}{[(-1)(x-1)]^3} - 5= \frac {2}{(-1)^3(x-1)^3} - 5 = \frac {2}{- (x-1)^3} - 5 = \frac {-2}{(x-1)^3} - 5\).

For a function like \(F(x) = \sqrt {4x-12} + 1 = \sqrt {4(x-6)} + 1 = \sqrt {4}\sqrt {x-3} + 1 = 2 \sqrt {x-3} + 1\), this approach works fine. However, if the coefficient of \(x\) is negative, for example, \(F(x) = \sqrt {1-x} = \sqrt {(-1)(x-1)}\) we get stuck the product rule for radicals doesn’t extend to negative quantities when the index is even. Hence we add an extra parameter which means we have an extra step. We state Theorem 1 below.

Proof. As usual, we ‘build’ the graph of \(F(x) = a \sqrt [n]{bx-h}+k\) starting with the graph of \(f(x) = \sqrt [n]{x}\) one step at a time. First, we consider the graph of \(F_{1}(x) = \sqrt {x-h}\). A generic point on the graph of \(F_{1}\) looks like \((x, \sqrt [n]{x-h})\). Note that if \(n\) is odd, \(x\) can be any real number whereas if \(n\) is even \(x-h \geq 0\) so \(x \geq h\). If we let \(c = x-h\), then \(x = c+h\) and we can change (dummy) variables and obtain a new representation of the point: \((c+h, \sqrt [n]{c})\). Note that if \(n\) is odd, \(x\) and \(c\) vary through all real numbers; if \(n\) is even, \(x \geq h\) and, hence, \(c \geq 0\). Since a generic point on the graph of \(f(x) = \sqrt [n]{x}\) can be represented as \((c, \sqrt [n]{c})\) for applicable values of \(c\), we see that we can obtain every point on the graph of \(F_{1}\) by adding \(h\) to each \(x\)-coordinate of the graph of \(f\), establishing step 1 of the theorem.

Proceeding to (the new!) step 2, a point on the graph of \(F_{2}(x) = \sqrt [n]{bx-h}\) has the form \((x, \sqrt [n]{bx-h})\). If \(n\) is odd, as usual, \(x\) can vary through all real numbers. If \(n\) is even, we require \(bx-h \geq 0\) or \(bx \geq h\). If \(b>0\), this gives \(x \geq \frac {h}{b}\). If, on the other hand, \(b<0\), the we have \(x \leq \frac {h}{b}\). Let \(c = bx\) and since by assumption \(b \neq 0\), we have \(x = \frac {c}{b}\). Once again, we change dummy variables from \(x\) to \(c\) and describe a generic point on the graph of \(F_{2}\) as \(\left ( \frac {c}{b}, \sqrt [n]{c - h} \right )\). If \(n\) is odd, \(x\) and \(c\) can vary through all real numbers. If \(n\) is even and \(b>0\), then \(x \geq \frac {h}{b}\) and, hence, \(c = bx \geq h\); if \(b<0\), then \(x \leq \frac {h}{b}\) also gives \(c = bx \geq h\). Since a generic point on the graph of \(F_{1}\) can be represented as \((c, \sqrt {c-h})\) for applicable values of \(c\), we see we can obtain every point on the graph of \(F_{2}\) by dividing every \(x\)-coordinate on the graph of \(F_{1}\) by \(b\), as per step 2 of the theorem.

The proof of steps 3 and 4 of Theorem 1 are identical to the proof of Theorem ?? (just with \(\sqrt [n]{\cdot }\) instead of \(( \cdot )^n\)) so we invite the reader to work through the details on their own. □

We demonstrate Theorem 1 in the following example.

2 Other Functions involving Radicals

Now that we have some practice with basic root functions, we turn our attention to more general functions involving radicals. In general, Calculus is the best tool with which to study these functions. Nevertheless, we will use what algebra we know in combination with a graphing utility to help us visualize these functions and preview concepts which are studied in greater depth in later courses. In the table below, we summarize some of the properties of radicals from elsewhere in this text (and Intermediate Algebra) we will be using in the coming examples.

We end this section with a classic application of root functions.