Section Preview
- Note to the Instructor (click to expand)
- The purpose of this very short section is to reintroduce students to rational exponents and to acquaint them with factoring expressions having rational exponents. These occur somewhat frequently in Calculus.
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
-
- Factor and simplify common expressions found in Calculus involving rational powers
While you will be presented with radical expressions in Calculus, more often than not these will be written with rational exponents. In fact, it is advantageous in Calculus to use rational exponents instead of radical notation.
When working with expressions in Calculus involving rational exponents, there are three skills we need to review (or introduce). In order of frequency and importance, these are:
- understanding the meaning and impact of a negative exponent,
- factoring out the GCF from an expression, and
- knowing how to multiply the numerator and denominator of an expression by a unit fraction to simplify.
The first of these is addressed with a quick reminder:\[a^{-n} = \dfrac{1}{a^n} \quad \text{and} \quad \dfrac{1}{a^{-n}} = a^n \nonumber\]We will be using this Law of Exponents frequently in this course and in Calculus.
The second might seem silly to mention (because you likely have been factoring out the GCF from algebraic expressions since elementary school); however, a simple change in the look of an expression throws many students into a downward spiral. The critical concept to remember when factoring out the GCF from an algebraic expression is that factoring is a division process. That is, when we factor \(3x - 6\), we divide out the \(3\) from both \(3x\) and \(-6\). Hence,\[3x - 6 = 3\left(\dfrac{3x}{3} - \dfrac{6}{3}\right) = 3 (x - 2). \nonumber\]Of course, most people skip that middle step. With this being said, we can focus on factoring more complex algebraic expressions.
When factoring the GCF out of an algebraic expression, we were taught to factor out the smallest-powered versions of the shared factors. For example, given \(a^M b^n + a^m b^N\), where \(m \lt M\) and \(n \lt N\), we identify the common factors as \(a\) (to a power) and \(b\) (to a power). Since the smallest power on \(a\) is \(m\) and the smallest power on \(b\) is \(n\), we factor out \(a^m b^n\).\[a^M b^n + a^m b^N = a^m b^n \left(\dfrac{a^M b^n}{a^m b^n} + \dfrac{a^m b^N}{a^m b^n}\right) = a^m b^n \left(a^{M-m} + b^{N - n}\right). \nonumber\]To memorize this formula is preposterous, so we will focus on the concept rather than the rote memorization.
Example \(\PageIndex{1}\)
Simplify the expression by factoring:\[3(1 + x)^{1/3} - x(1 + x)^{-2/3} \nonumber\]
- Solution
-
Both terms in the expression share the factor \((1 + x)\); however, the second term has the smallest-powered version of this factor. Therefore, we factor out \(\left(1 + x\right)^{-2/3}\).\[\begin{array}{rclcl}
3(1 + x)^{1/3} - x(1 + x)^{-2/3} & = & (1 + x)^{-2/3} \left(\dfrac{3(1 + x)^{1/3}}{(1 + x)^{-2/3}} - \dfrac{x (1 + x)^{-2/3}}{(1 + x)^{-2/3}}\right) & \quad & \left( \text{factoring out the GCF (which is just division)} \right) \\[6pt]
& = & (1 + x)^{-2/3} \left(3(1 + x)^{1/3 + 2/3} - x (1 + x)^{-2/3 + 2/3}\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & (1 + x)^{-2/3} \left(3(1 + x) - x (1 + x)^{0}\right) & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & (1 + x)^{-2/3} \left(3(1 + x) - x\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & (1 + x)^{-2/3} \left(3 + 2x\right) & \quad & \left( \text{distributing and combining like terms} \right) \\[6pt]
\end{array} \nonumber\]
If you pay close attention, Example \(\PageIndex{1}\) reveals a shortcut to factoring out the smallest-powered version of each shared factor: factoring out \((1 + x)^{-2/3}\) ends up subtracting \(-\frac{2}{3}\) from the remaining exponents. That is, it adds the \(\frac{2}{3}\) to those exponents. If you pay attention and do a few examples of your own, this trick speeds up factoring problems like this significantly.
Another thing to note from Example \(\PageIndex{1}\) is that we left our answer with a negative exponent—this is perfectly fine (and actually preferred in Calculus).
The next example demonstrates that we always try to factor completely, if possible.
Example \(\PageIndex{2}\)
Factor:\[x\left(2x + 7\right)^{1/3} + 5 \left(2x + 7\right)^{-2/3} \nonumber\]
- Solution
- \[\begin{array}{rclcl}
x\left(2x + 7\right)^{1/3} + 5 \left(2x + 7\right)^{-2/3} & = & \left(2x + 7\right)^{-2/3} \left(x \left(2x + 7\right)^{3/3} + 5 \left(2x + 7\right)^0\right) & \quad & \left( \text{factoring out the GCF} \right) \\[6pt]
& = & \left(2x + 7\right)^{-2/3} \left(x \left(2x + 7\right) + 5\right) & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \left(2x + 7\right)^{-2/3} \left(2x^2 + 7x + 5\right) & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \left(2x + 7\right)^{-2/3} \left(2x + 5\right)\left(x + 1\right) & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\end{array} \nonumber\]
Checkpoint \(\PageIndex{2}\)
Expressions like those in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) commonly occur as the result of the Product Rule in Differential Calculus (Calculus I). You will have need to simplify those without any prompting from your instructor.
Another style of algebraic expression that frequently occurs in Calculus is in the following example.
Example \(\PageIndex{3}\)
Simplify the expression.\[\dfrac{2\left(18 + x\right)^{1/2} - x\left(18 + x\right)^{-1/2}}{x + 18} \nonumber\]
- Solution
- There are two approaches to simplifying expressions like these. The first is to mimic what we did in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\).
Method #1\[\begin{array}{rclcl}
\dfrac{2\left(18 + x\right)^{1/2} - x\left(18 + x\right)^{-1/2}}{x + 18} & = & \dfrac{\left(18 + x\right)^{-1/2} \left(2\left(18 + x\right)^{2/2} - x\left(18 + x\right)^{0}\right)}{x + 18} & \quad & \left( \text{factoring the GCF from the numerator} \right) \\[6pt]
& = & \dfrac{\left(18 + x\right)^{-1/2} \left(2\left(18 + x\right) - x\right)}{x + 18} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{\left(18 + x\right)^{-1/2} \left(36 + 2x - x\right)}{x + 18} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{\left(18 + x\right)^{-1/2} \left(36 + x\right)}{x + 18} & \quad & \left( \text{combining like terms} \right) \\[6pt]
& = & \dfrac{\left(36 + x\right)}{\left(18 + x\right)^{1/2} \left(x + 18\right)} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & \dfrac{36 + x}{\left(18 + x\right)^{3/2}} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\end{array} \nonumber\]While this method works, and there is nothing wrong with it, it is helpful to show another approach.
Method #2
This approach will use the fact that \(a^{-n}\) is the fraction \(\frac{1}{a^n}\).\[\begin{array}{rclcl}
\dfrac{2\left(18 + x\right)^{1/2} - x\left(18 + x\right)^{-1/2}}{x + 18} & = & \dfrac{2\left(18 + x\right)^{1/2} - x \frac{1}{\left(18 + x\right)^{1/2}}}{x + 18} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & \dfrac{\left(2\left(18 + x\right)^{1/2} - x \frac{1}{\left(18 + x\right)^{1/2}}\right)}{\left(x + 18\right)} \cdot \dfrac{\left(18 + x\right)^{1/2}}{\left(18 + x\right)^{1/2}} & \quad & \left( \text{simplifying the compound fraction} \right) \\[6pt]
& = & \dfrac{2\left(18 + x\right)^{2/2} - x}{\left(x + 18\right)^{3/2}} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{2\left(18 + x\right) - x }{\left(x + 18\right)^{3/2}} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{36 + 2x - x }{\left(x + 18\right)^{3/2}} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{36 + x}{\left(x + 18\right)^{3/2}} & \quad & \left( \text{combining like terms} \right) \\[6pt]
\end{array} \nonumber\]
While both approaches in Example \(\PageIndex{3}\) look like they take the same amount of work, the second (Method #2) is often a little faster. Expressions of the form in Example \(\PageIndex{3}\) are the result of the Quotient Rule in Differential Calculus. As before, you will have need to simplify these without any prompting from your instructor.
Checkpoint \(\PageIndex{3}\)
Success in College: Learning From Tests (Part 2)
- A student earned 75% of the points on her last math test. On the first class day after the test, the class started a new chapter with new material. She does not understand several vital procedures from the last test. What do you think she should do about this situation?
- A student usually finishes tests right as class ends. He doesn't have time to check for careless mistakes. He looks carefully at his corrected tests. He realizes that many of his errors are arithmetic or algebra mistakes. He knew how to do the problems; he didn't do them well enough. What do you think he should do?
Miscellaneous Topics: Thinking About Mathematics (Part 2)
A significant concern in mathematics education is the trend that students rely on calculators more. An argument for calculator usage is that, in "real life," a person will always have access to a computer or a calculator. Therefore, forcing students to do problems "by hand" is useless.
- Give an argument against high calculator usage in math classes.
- Do you think that your professor relies too much on calculator usage in class? What about their use of technology? Would you like more or less?
