]> Chapter 07 Notes

7: Radical Functions and Rational Exponents

7-1: Roots and Radical Expressions

If $a^{n} = b$ , then a is the n^th root of b. Note that this is a different kind of root than the one that we talked about in Chapter 6—here, we talk about roots of numbers; previously, we talked about roots of equations.

The notation for the n^th root of b is $\sqrt[n]{b}$ . n is called the index; b is called the radicand. The radix (√) was first used by German mathematician Christoff Rudolff in 1525. The horizontal bar (the virgule) was added by French mathematician Rene Descartes in 1637.

If the index is odd, then there is one real n^th root of a real number.

If the index is even and the radicand is non-negative, then there are two real n^th roots of the radicand. In this case, the two numbers will have equal absolute values and opposite signs. The principal root is the one that is positive. If no sign is indicated in front of a radical (if the radical is given), then the expression is the principal root.

(If you allow complex numbers, then can you guess how many complex n^th roots there are for a complex number?)

Important note: if n is even, then $\sqrt[n]{a^{n}} = | a |$ . Since there is no sign in front of the radical, we are required to give the principal root.

We will see how to get the calculator to produce roots (other than square roots) in class.

You already know how to simplify radical expressions, so there is no need to repeat it here.

Examples

[1.] Find all square roots of 45.

$\pm \sqrt{45} = \pm 3 \sqrt{5} \approx \pm 6.708$

[2.] Simplify $\sqrt[4]{81 a^{4} b^{5}}$ .

$\sqrt[4]{81 a^{4} b^{5}} = \sqrt[4]{81} \cdot \sqrt[4]{a^{4} b^{4}} \cdot \sqrt[4]{b} = 3 a b \sqrt[4]{b}$

[3.] Solve $x^{3} = 64$ .

$x^{3} = 64 ⇒ x = \sqrt[3]{64} = 4$ . Since the index is odd, there is only one answer.

7-2: Multiplying and Dividing Rational Expressions

You are allowed to multiply two radical expressions if the indices are the same—square roots can be multiplied with square roots; cube roots can be multiplied with cube roots; etc.

The same is true for division.

A related skill is rationalizing the denominator. This means removing all radical expressions from the denominator of a fraction. To do this, multiply the existing fraction by one—a one created with the radical part of the denominator.

A long time ago (prior to the 1990's), before calculators really came around, rationalizing the denominator was an important skill. Now, the most important reason to do it is to earn all of the points on your test when someone asks you to rationalize…

Examples

[4.] Multiply and simplify: $\sqrt[3]{50 x^{2} z^{5}} \cdot \sqrt[3]{15 y^{3} z}$ .

$\sqrt[3]{50 x^{2} z^{5}} \cdot \sqrt[3]{15 y^{3} z} = \sqrt[3]{750 x^{2} y^{3} z^{6}} = \sqrt[3]{125 y^{3} z^{6}} \cdot \sqrt[3]{6 x^{2}} = 5 y z^{2} \sqrt[3]{6 x^{2}}$

[5.] Divide and simplify: $\frac{\sqrt[4]{243 k^{3}}}{\sqrt[4]{3 k^{7}}}$ .

$\frac{\sqrt[4]{243 k^{3}}}{\sqrt[4]{3 k^{7}}} = \sqrt[4]{\frac{243 k^{3}}{3 k^{7}}} = \sqrt[4]{81 k^{4}} = 3 k$

[6.] Rationalize the denominator, assuming that all variables are positive: $\frac{\sqrt{x y}}{\sqrt{3 x}}$ .

$\frac{\sqrt{x y}}{\sqrt{3 x}} \cdot \frac{\sqrt{3 x}}{\sqrt{3 x}} = \frac{\sqrt{3 x^{2} y}}{\sqrt{9 x^{2}}} = \frac{x \sqrt{3 y}}{3 x} = \frac{\sqrt{3 y}}{3}$

7-3: Binomial Radical Expressions

Radical expressions can be added and subtracted only if they have the same index and same radicand. When this is true, only the coefficient in front of the radical changes.

A binomial radical expression is a number added or subtracted with a radical. Multiplying binomial radical expressions simply requires the FOIL method. Rationalizing the denominator when it is a binomial radical expression requires that you use the conjugate. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$ .

Examples

[7.] Multiply and simplify: $(2 + 3 \sqrt{7}) (3 - 4 \sqrt{7})$ .

$(2 + 3 \sqrt{7}) (3 - 4 \sqrt{7}) = (2) (3) + (2) (- 4 \sqrt{7}) + (3 \sqrt{7}) (3) + (3 \sqrt{7}) (- 4 \sqrt{7}) = 6 - 8 \sqrt{7} + 9 \sqrt{7} - 12 \sqrt{49} = 6 + \sqrt{7} - 12 \cdot 7 = - 78 + \sqrt{7}$

[8.] Rationalize the denominator: $\frac{1 - \sqrt{2}}{3 + \sqrt{5}}$ .

$\frac{1 - \sqrt{2}}{3 + \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{3 - \sqrt{5} - 3 \sqrt{2} + \sqrt{10}}{9 - \sqrt{25}} = \frac{3 - \sqrt{5} - 3 \sqrt{2} + \sqrt{10}}{4}$

7-4: Rational Exponents

Another way to indicate radicals is to use exponents that are fractions. The first modern use of rational exponents is credited to Isaac Newton (1643-1727), an English mathematician and scientist, co-inventor of The Calculus.

The notation: $\sqrt[n]{x} = x^{\frac{1}{n}}$ . When written in this form, radicals obey the same rules as other exponents. These rules are summarized for you in your textbook…

Examples

[9.] Simplify: ${(x^{\frac{2}{3}}, y^{\frac{3}{4}})}^{12}$ .

${(x^{\frac{2}{3}}, y^{\frac{3}{4}})}^{12} = x^{12 \cdot \frac{2}{3}} \cdot y^{12 \cdot \frac{3}{4}} = x^{8} y^{9}$

[10.] Rewrite using radicals and positive exponents: $z^{- \frac{5}{2}}$ .

$z^{- \frac{5}{2}} = \frac{1}{z^{\frac{5}{2}}} = \frac{1}{\sqrt{z^{5}}}$

7-5: Solving Radical Equations

If you can isolate the variable, then do so. Raise each side to the required power so that the variable is now to the first power. Continue solving as necessary.

If you can't isolate the variable, then isolate one instance, and raise both sides of the equation to the necessary power, so that the power of the variable (the one instance that you isolated) is one. Now combine like terms, and work on isolating the variable again…note that this may mean that you have to raise both sides of the equation to another power!

This is difficult to say in words…watch our examples in class closely.

When raising expressions to powers, you may accidentally introduce extraneous solutions—solutions that work in later stages of the equation, but do not work in the original equation. Thus, it will now be important that you check your solutions.

Examples

[11.] Solve $7 + \sqrt{2 x - 1} = 10$ .

$7 + \sqrt{2 x - 1} = 10 ⇒ \sqrt{2 x - 1} = 3 ⇒ 2 x - 1 = 9 ⇒ 2 x = 10 ⇒ x = 5$ . Now, check: $7 + \sqrt{2 (5) - 1} = 7 + \sqrt{10 - 1} = 7 + \sqrt{9} = 7 + 3 = 10$ . It works; the solution is $x = 5$ .

[12.] Solve $\sqrt{x + 7} = x - 5$ .

$\sqrt{x + 7} = x - 5 ⇒ x + 7 = {(x - 5)}^{2} ⇒ x + 7 = x^{2} - 10 x + 25 ⇒ 0 = x^{2} - 11 x + 18$ . Factor this to get $0 = (x - 9) (x - 2)$ , which indicates that either $x = 9$ or $x = 2$ . Check: $\sqrt{9 + 7} \overset{?}{=} 9 - 5 ⇒ \sqrt{16} \overset{?}{=} 4 ⇒ 4 = 4$ . $\sqrt{2 + 7} \overset{?}{=} 2 - 5 ⇒ \sqrt{9} \overset{?}{=} - 3 ⇒ 3 \neq - 3$ . The only solution is $x = 9$ .

7-6: Function Operations

In several previous instances, I've alluded to the idea that functions can be treated much like numbers…let's pull all of that together with some notation.

$(f \pm g) (x) = f (x) \pm g (x)$

$(f g) (x) = f (x) \cdot g (x)$

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}$

There is one more way to combine functions—called composition of functions. This is where one function is put inside of another function. This is an extremely important idea!

$(f ◦ g) (x) = f (g (x))$

Examples

[13.] If $f (x) = x^{2} + 2 x$ and $g (x) = 3 x - 5$ , then find $(f + g) (x)$ , $(f - g) (x)$ , $(f g) (x)$ , $(\frac{f}{g}) (x)$ , $(f ◦ g) (x)$ , and $(g ◦ f) (x)$ .

$(f + g) (x) = x^{2} + 2 x + 3 x - 5 = x^{2} + 5 x - 5$

$(f - g) (x) = (x^{2} + 2 x) - (3 x - 5) = x^{2} + 2 x - 3 x + 5 = x^{2} - x + 5$

$(f g) (x) = (x^{2} + 2 x) (3 x - 5) = 3 x^{3} - 5 x^{2} + 6 x^{2} - 10 x = 3 x^{3} + x^{2} - 10 x$

$(\frac{f}{g}) (x) = \frac{x^{2} + 2 x}{3 x - 5}$ , which really can't be simplified.

$(f ◦ g) (x) = {(3 x - 5)}^{2} + 2 (3 x - 5) = 9 x^{2} - 30 x + 25 + 6 x - 10 = 9 x^{2} - 24 x + 15$

$(g ◦ f) (x) = 3 (x^{2} + 2 x) - 5 = 3 x^{2} + 6 x - 5$

[14.] If $x (t) = 2 t + 1$ and $y (t) = \sqrt{t}$ , then find $(x ◦ y) (9)$ .

$(x ◦ y) (t) = 2 \sqrt{t} + 1$ , so $(x ◦ y) (9) = 2 \sqrt{9} + 1 = 6 + 1 = 7$ .

7-7: Inverse Relations and Functions

Recall that any set of points in the plane—and thus, any equation—represents a relation.

The inverse relation is the reflection of a relation across the line $y = x$ . This reflection has the effect of swapping x and y. Thus, if your relation has an equation, the way to find the inverse relation is to swap x and y. Note that this swapping means that the domain of one relation will be the range of the inverse relation, and vice versa.

Since every function is also a relation, every function has an inverse relation. The question is: are there functions whose inverse relations are also functions?

Think logically: if a relation is a function, then it obeys the vertical line test—every line of the form $x = a$ intersects the relation at exactly one point. Now, flip that across the line $y = x$ …where we had a vertical line crossing the relation at one place, we now have a horizontal line crossing the relation at one point. Thus, if the inverse of a relation is a function, the relation must pass the horizontal line test—every horizontal line crosses the function at one point.

So: to be a function, the relation must pass the vertical line test, and in order to have an inverse function, the relation (function) must pass the horizontal line test.

A relation that passes both the vertical and horizontal line tests is called a one-to-one function. For every x in the domain, there is one and only one y in the range.

A note: if you compose a function with its inverse function, then everything except the input (usually x) cancels out.

The horizontal line test is only useful if you plan on graphing, or have a graph of the function. If you don't, swap x and y, then solve for y. If you can solve for y without a ±, then you've probably got an inverse function—the only thing you may still need to check is that the domain and range have exactly swapped.

Examples

[15.] Find the inverse relation of $y = 2 x - 5$ . Is the inverse relation a function?

The inverse relation is $x = 2 y - 5 ⇒ x + 5 = 2 y ⇒ \frac{1}{2} x + \frac{5}{2} = y$ . I know that this is an inverse function because the graphs of both things are lines, which (with two particular exceptions) pass both the vertical and horizontal line tests.

[16.] Find the inverse relation of $y = {(x - 2)}^{3} + 9$ . Is the inverse relation a function?

The inverse relation is $x = {(y - 2)}^{3} + 9 ⇒ x - 9 = {(y - 2)}^{3} ⇒ \sqrt[3]{x - 9} = y - 2 ⇒ 2 + \sqrt[3]{x - 9} = y$ . Since I didn't end up with a ±, I'm pretty sure that this is an inverse function (check the domain and range; it is).

[17.] If $h (y) = 6 y - 13$ , then find the value of $(h^{-1} ◦ h) (π)$ .

Since $h (y)$ is linear (and not one of the two exceptions), it has an inverse. Thus, $(h^{-1} ◦ h) (y) = y$ . Hence, $(h^{-1} ◦ h) (π) = π$ .

7-8: Graphing Radical Functions

Once again, you want to get away from the "pick some x-values, plot the points" approach. Instead, memorize a couple of key points and shapes, then use the ultra handy-dandy templates.

The main difference is in whether the index is even or odd. If the index is even, then the parent graph will have a domain of $x \geq 0$ , and will pass through the points $(0, 0)$ , $(1, 1)$ and $(2^{n}, 2)$ . Here are a couple of examples (indices 2, 4 and 6):

An Image

If the index is odd, then the domain will be all real numbers, and the parent graph will pass through the points $(0, 0)$ , $(\pm 1, 1)$ and $(\pm 2^{n}, 2)$ . Here are a few more examples (indices 3, 5 and 7):

An Image

The template looks like this: $y = a \sqrt[n]{b (x - c)} + d$ . Remember that a multiplies the y-values, b divides the x-values, c moves the graph left or right, and d moves the graph up and down.

Examples

Well…you can check these with your calculator, so you shouldn't need any examples here.

Page last validated 2010-08-15