2.1 Difference Quotients of Polynomial Functions

Purpose. To evaluate and simplify the difference quotient of polynomials of any degree, and to apply it to evaluate the difference quotients of rational functions and functions involving radicals.
 

Prerequisites. It is expected that you have studied 1.1.7 and completed all problems in 1.1.8 before beginning your study of this section.

2.1.1 Difference Quotients

 
If \(x\) and \(a\) are numbers in the domain of a function \(y=f(x)\) with \(x\neq a\), then the points \(\left(x,f\left(x\right)\right)\) and \(\left(a,f\left(a\right)\right)\) are on the graph of \(f\), and the \textbf{slope of the secant line} to the graph passing through \(\left(x,f\left(x\right)\right)\) and \(\left(a,f\left(a\right)\right)\) is given by
\begin{equation}\label{ch02-eq01}
m=\frac{f\left(x\right)-f\left(a\right)}{x-a}\tag{2.1}.
\end{equation}

By introducing the \textit{change of variable} \(h = x-a\), so that \(x = a + h\), the above equation can be rewritten as
\begin{equation}\label{ch02-eq02}
m=\frac{f\left(a+h\right)-f\left(a\right)}{h}\tag{2.2}
\end{equation}
These two formulas are called the difference quotient or average rate of change of \(f\) between \(a\) and \(x\) (or \(x\) and \(x+h\)).

The pictures below illustrate this in the case of a quadratic function \(f\).

 

2.1.2 Factorization of \(a^n-b^n\) and Difference Quotients

 

Given \(f(x)= x^5-4x^3+2x^2-1\), we can compute its difference quotient term by term as follows:

\begin{align*}
\frac{f(x)-f(a)}{x-a} &= \frac{\left (x^5-4x^3+2x^2-1\right ) – \left (a^5-4a^3+2a^2-1\right )}{x-a} =\\
&= \frac{x^5-a^5}{x-a} – \frac{4 \left (x^3-a^3\right )}{x-a} + \frac{2(x^2-a^2)}{x-a}=\\
&= \frac{x^5-a^5}{x-a} – \frac{4(x-a)\left (x^2+xa + a^2\right )}{x-a} + \frac{2(x-a)(x+a)}{x-a}=\\
&= \frac{x^5-a^5}{x-a} – 4 \left (x^2+xa + a^2\right ) + 2(x+a).
\end{align*}

The issue is that so far, we do not have a factorization formula for \(x^5-a^5\). In general, we would like to factor the binomial \(x^n-a^n\) for \(n\in \mathbb{N}\), \(n \geq 2\). There is a very simple pattern to accomplish this.

\begin{align*}
a\hphantom{^2} – b\hphantom{^2} &= (a – b)\\
a^2 – b^2 &= (a – b)\left(a + b\right)\\
a^3 – b^3 &= (a – b)\left(a^2 + ab + b^2\right)\\
a^4 – b^4 &= (a – b)\left(a^3 + a^2b + a b^2+ b^3\right)\\
a^5 – b^5 &= (a – b)\left(a^4 + a^3b + a^2 b^2+ ab^3+b^4\right)
\end{align*}
On Your Own 1. (a) Write the next two rows of the pattern; (b) propose a general rule for any \(n \in \mathbb{N}\).

Proposition 1. Let \(a, b \in \mathbb{R}\) and \(n\in \mathbb{N}\). Then
\begin{equation}\label{ch02-eq03}
a^n – b^n = (a – b)\left(a^{n-1} + a^{n-2}b + a^{n-3} b^2+ \cdots + a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)\tag{2.3}.
\end{equation}
In particular, for \(x\neq a\),
\begin{equation}\label{ch02-eq04}
\frac{x^n-a^n}{x-a}=x^{n-1} + x^{n-2}a + x^{n-3}a^2 +\cdots x^2a^{n-3} +x a^{n-2}+ a^{n-1}\tag{2.4},.
\end{equation}

Proof. Let \(S\) denote the sum on the right hand side in \ref{ch02-eq03}. Then, using the distributive property and subtracting, we have
\[\renewcommand{\arraystretch}{2}
\begin{array}{rcl}
aS &= &a^{n} + a^{n-1}b + a^{n-2} b^2+ \cdots + a^3b^{n-3}+a^2b^{n-2}+ab^{n-1} \\
bS& = &a^{n-1}b + a^{n-2}b^2 + a^{n-3} b^3+ \cdots + a^2b^{n-2}+ab^{n-1}+b^{n} \vphantom{\displaystyle \frac{A^A}{A^{A^A}}_{A_A}} \\ \hline
aS-bS& = & a^{n} -b^{n} \vphantom{\displaystyle \frac{A^A}{A^{A^A}}_{A_A}}
\end{array}
\]
Then (\ref{ch02-eq03}) and (\ref{ch02-eq04}) follow immediately.

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Corollary 1. Let \(a, b \in \mathbb{R}\) and \(n\in \mathbb{N}\) with \(n\) odd. Then
\begin{equation}\label{ch02-eq05}
a^n + b^n = (a + b)\left(a^{n-1} – a^{n-2}b + a^{n-3} b^2- \cdots + a^2b^{n-3}-ab^{n-2}+b^{n-1}\right)\tag{2.5}.
\end{equation}

Proof. This is immediate and is based on the fact that \((-b)^n=-b^n\), for \(n\) odd, so that
\begin{align*}
a^n + b^n &= a^n – (-b)^n =\\
&=
(a -(- b))\left(a^{n-1} + a^{n-2}(-b) + a^{n-3} (-b)^2+ \cdots + a^2 (-b)^{n-3}-a (-b)^{n-2}+(-b)^{n-1}\right) \vphantom{\displaystyle \frac{A^A}{A_{A_A}}}\\
&= (a + b)\left(a^{n-1} – a^{n-2}b + a^{n-3} b^2- \cdots + a^2b^{n-3}-ab^{n-2}+b^{n-1}\right).
\end{align*}

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Corollary 2. Let \(x,\, a\in \mathbb{R}\), \(x\neq a\), and \(x a\neq 0\). Then
\begin{equation}\label{ch02-eq06}
\frac{x^{-n}-a^{-n}}{x-a}= -\frac{1}{x^na^n}\left (x^{n-1} + x^{n-2}a + x^{n-3}a^2 +\cdots x^2a^{n-3} +x a^{n-2}+ a^{n-1}\right )\tag{2.6}.
\end{equation}

Proof.
Let \(x\neq a\), then
\begin{align*}
\frac{x^{-n}-a^{-n}}{x-a}& =\frac{\displaystyle \frac{1}{x^n} – \frac{1}{a^n}}{x-a}
= \frac{ a^n-x^n}{(x-a)x^na^n}
= -\frac{1}{x^na^n}\frac{ x^n-a^n}{x-a} =\\
& \\
&\underset{(\ref{ch02-eq04})}{=} -\frac{1}{x^na^n}\left (x^{n-1} + x^{n-2}a + x^{n-3}a^2 +\cdots x^2a^{n-3} +x a^{n-2}+ a^{n-1}\right ).
\end{align*}

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Example 1. Given the function \(f(x)=2x^7-3x^6+5\), find and simplify its difference quotient (a) as in (\ref{ch02-eq01}); (b) as in (\ref{ch02-eq02}).

Solution. (a) for \(x \neq a\),
\[\frac{f\left(x\right)-f\left(a\right)}{x-a}=
\frac{(2x^7-3x^6+5)-\left(2a^7-3a^6+5\right)}{x-a}=
\frac{2(x^7-a^7)}{x-a}-\frac{3(x^6-a^6)}{x-a}
=\]
\[=2(x^6 + x^5a + x^4a^2 + x^3a^3 +x^2 a^4+ xa^5 + a^6) –
3(x^5 + x^4a + x^3a^2 + x^2a^3 +x a^4+ a^5)\,.
\]

(b) Similarly, for \(h \neq 0\), we have,
\begin{align*}
&\frac{f(a+h)-f(a)}{h}
=
2\frac{(a+h)^7-a^7}{h}-3\frac{(a+h)^6-a^6}{h}+\frac{5-5}{h}=\\
&=2((a+h)^6 + (a+h)^5a + (a+h)^4a^2 + (a+h)^3a^3 +(a+h)^2a^4+(a+h) a^5+a^6)\vphantom{\displaystyle \frac{A^A}{A_A}}\\
&\hspace{.5in}
-3((a+h)^5 + (a+h)^4a + (a+h)^3a^2 + (a+h)^2a^3 +(a+h)a^4+a^5)
\,.
\end{align*}
Note that in (b), it is more efficient to factor the numerator first, rather than expanding the terms \((a+h)^7\) and \((a+h)^6\), to cancel the common factor \(h\).

Example 2. Let \(\displaystyle f(x) = \frac{3}{x^5}- \frac{4}{x^3}+ 2x^2\). Find and simplify the difference quotient \(\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}\).

Solution. Assume that \(x\neq a\), then
\begin{align*}
\frac{f\left(x\right)-f\left(a\right)}{x-a}&= \frac{\displaystyle \frac{3}{x^5}- \frac{4}{x^3}+ 2x^2-\left(\frac{3}{a^5}- \frac{4}{a^3}+ 2a^2\right)}{x-a}
= \frac{\displaystyle \frac{3}{x^5}- \frac{3}{a^5}- \frac{4}{x^3}+ \frac{4}{a^3}+ 2x^2 – 2a^2 }{x-a} =\\
& \\
& = \frac{3\left (a^5-x^5\right )}{x^5a^5(x-a)} – \frac{4\left (a^3-x^3\right )}{x^3a^3(x-a)}+ \frac{2\left (x^2 – a^2 \right )}{x-a} =\\
& \\
&= \frac{-3\left (x^4 + x^3a + x^2a^2+xa^3+a^4\right )}{x^5a^5} +
\frac{4\left (x^2+ xa + a^2\right )}{x^3a^3}+ 2\left (x +a \right ).
\end{align*}

2.1.3 Difference Quotients of Roots

 

Example 1. Given \(\displaystyle f(x)=\sqrt[5]{x}\), find and simplify the difference quotient \(\displaystyle \frac{f(x)-f(a)}{x-a}\).

Solution. We can proceed in two different ways—cf. Example 4 in 1.1.7.
\begin{align*}
\frac{\sqrt[5]{x} – \sqrt[5]{a}}{x-a} &=
\frac{\sqrt[5]{x} – \sqrt[5]{a}}{x-a} \cdot\frac{\sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4}}{\sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4}} =\\
&=\frac{x-a}{(x-a)\left (\sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4}\right )}=\\
&=\frac{1}{\sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4}}, \;\; x\neq a.
\end{align*}
Alternatively
\begin{align*}
\frac{\sqrt[5]{x} – \sqrt[5]{a}}{x-a} &=\frac{\sqrt[5]{x} – \sqrt[5]{a}}{\left (\sqrt[5]{x}\right )^5 – \left (\sqrt[5]{a}\right )^5} = \frac{\sqrt[5]{x} – \sqrt[5]{a}}{\left ( \sqrt[5]{x} – \sqrt[5]{a} \right )\left ( \sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4} \right ) } =\\
&= \frac{1}{\sqrt[5]{x^4}+\sqrt[5]{x^3a} +\sqrt[5]{x^2 a^2}+\sqrt[5]{x a^3}+\sqrt[5]{a^4}}, \;\; x\neq a.
\end{align*}

Example 2. Given \(\displaystyle f(x)=\frac{1}{\sqrt[5]{x}}\), find and simplify the difference quotient \(\displaystyle \frac{f(x)-f(a)}{x-a}\).

Solution.
\begin{align*}
\frac{\displaystyle \frac{1}{\sqrt[5]{x}} – \frac{1}{\sqrt[5]{a}} }{x-a} &=
-\frac{\sqrt[5]{x} – \sqrt[5]{a}}{(x-a)\,\sqrt[5]{x}\,\sqrt[5]{a} } = \frac{1}{\sqrt[5]{x a}} \cdot \frac{1}{\sqrt[5]{x^4}+\sqrt[5]{x^3a} +\sqrt[5]{x^2 a^2}+\sqrt[5]{x a^3}+\sqrt[5]{a^4}}, \;\; x\neq a,
\end{align*}
where we have used Example 1.

Here is an example that is slightly different.

Example 3. Given \(\displaystyle f(x)=\sqrt[5]{x^3}\), find and simplify the difference quotient \(\displaystyle \frac{f(x)-f(a)}{x-a}\).

Solution.
\begin{align*}
\frac{\sqrt[5]{x^3} – \sqrt[5]{a^3}}{x-a} &=
\frac{\left (\sqrt[5]{x}\,\right )^3 – \left (\sqrt[5]{a}\,\right )^3}{x-a} =
\frac{\left (\sqrt[5]{x} – \sqrt[5]{a}\,\right )\left (\left (\sqrt[5]{x}\,\right )^2 +\sqrt[5]{x} \sqrt[5]{a} + \left (\sqrt[5]{a}\,\right )^2\right )}{x-a} =\\
&= \frac{\sqrt[5]{x} – \sqrt[5]{a}}{x-a} \cdot\left (\sqrt[5]{x^2} +\sqrt[5]{xa} + \sqrt[5]{a^2}\right ).
\end{align*}
And, using Example 1 above, we have
\[\frac{\sqrt[5]{x^3} – \sqrt[5]{a^3}}{x-a} =\frac{\sqrt[5]{x^2} +\sqrt[5]{xa} + \sqrt[5]{a^2}}{\sqrt[5]{x^4}+\sqrt[5]{x^3}\,\sqrt[5]{a}+\sqrt[5]{x^2}\,\sqrt[5]{a^2}+\sqrt[5]{x}\,\sqrt[5]{a^3}+\sqrt[5]{a^4}}, \;\; x\neq a.\]

Corollary 3. Let \(x,\, a\in \mathbb{R}\), \(x\neq a\). Then
\begin{equation}\label{ch02-eq07}
\frac{\sqrt[n]{x}-\sqrt[n]{a}}{x-a}= \frac{1}{\sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} a} + \cdots \sqrt[n]{x^2a^{n-3}} +\sqrt[n]{x a^{n-2}}+\sqrt[n]{a^{n-1}}}.
\end{equation}
\begin{equation}\label{ch02-eq08}
\frac{\displaystyle \frac{1}{\sqrt[n]{x}}-\frac{1}{\sqrt[n]{a}}}{x-a}= -\frac{1}{\sqrt[n]{x a}}\cdot\frac{1}{\sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} a} + \cdots \sqrt[n]{x^2a^{n-3}} +\sqrt[n]{x a^{n-2}}+\sqrt[n]{a^{n-1}}}.
\end{equation}

\begin{equation}\label{ch02-eq09}
\frac{\sqrt[n]{x^m}-\sqrt[n]{a^m}}{x-a}= \frac{\sqrt[n]{x^{m-1}}+\sqrt[n]{x^{m-2} a} + \cdots \sqrt[n]{x^2a^{m-3}} +\sqrt[n]{x a^{m-2}}+\sqrt[n]{a^{m-1}}}{\sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} a} + \cdots \sqrt[n]{x^2a^{n-3}} +\sqrt[n]{x a^{n-2}}+\sqrt[n]{a^{n-1}}}.
\end{equation}

Proof. (a)
Let \(x\neq a\), then
\begin{align*}
\frac{\sqrt[n]{x}-\sqrt[n]{a}}{x-a}& =\frac{\sqrt[n]{x}-\sqrt[n]{a}}{\left (\sqrt[n]{x}\right )^n- \left (\sqrt[n]{a}\right )^n} = \\
&=\frac{\sqrt[n]{x}-\sqrt[n]{a}}{\left (\sqrt[n]{x} – \sqrt[n]{a}\right )\cdot \left ( \sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} a} + \cdots \sqrt[n]{x^2a^{n-3}} +\sqrt[n]{x a^{n-2}}+\sqrt[n]{a^{n-1}} \right )} =\\
&=\frac{1}{ \sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} a} + \cdots \sqrt[n]{x^2a^{n-3}} +\sqrt[n]{x a^{n-2}}+\sqrt[n]{a^{n-1}}}.
\end{align*}
(b) The proof is similar to Example 2 and is left to the reader to complete the details.

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