Page 3.1: Difference Quotient
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A big part of calculus is being able to understand the notation when it comes to dealing with functions. For this reason, we will spend some time on discussing how to think about our notation.
Suppose we have the function \( f: \mathbb{R} \rightarrow \mathbb{R} \) specified by \(f(x) = x^2 \). To recap a few things, this notation is telling us that the domain is \( \mathbb{R} \) meaning that the \(x\) values can be any real number. When we plug in these \(x\) values into \(f\), the outputs are also real numbers and hence our codomain is \( \mathbb{R} \).
Allow us to consider the following exercise.
Suppose \( f: \mathbb{R} \rightarrow \mathbb{R} \) specified by \(f(x) = x^2 \). Find the following:
1) \( f(1) \)
2) \( f(2) \)
3) \( f(3) \)
4) \( f(a) \)
5) \( f(b) \)
6) \( f(a+h) \)
7) \( \frac{f(a+h) - f(a)}{h} \)
- Answer
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1) The notation \(f(1)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by a 1. Doing so yields \( f(1) = 1^2 = 1 \)
2) The notation \(f(2)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by a 2. Doing so yields \( f(2) = 2^2 = 4 \)
3) The notation \(f(3)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by a 3. Doing so yields \( f(3) = 3^2 = 9\)
4) The notation \(f(a)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by an \(a\). Doing so yields \(f(a) = a^2 \).
5) The notation \(f(b)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by a \(b\). Doing so yields \(f(b) = b^2 \).
6) The notation \( f(a+h)\) is telling us that wherever we see \(x\) in the function \(f(x) = x^2 \), we should replace the \(x\) by an \( a+h\). Doing so yields \(f(a+h) = (a+h)^2 = a^2 + 2ah + h^2 \).
7) Putting Parts 4 and 6 together yields: \begin{align*} \frac{f(a+h) - f(a)}{h} &= \frac{(a+h)^2 - a^2}{h } \\ &= \frac{a^2 + 2ah + h^2 - a^2}{h} \\&= \frac{2ah + h^2}{h} \\ &= 2a + h \end{align*}
The quantity we considered in Part 7 has a special name - the difference quotient.
Definition: Given a function \(f\), we refer to \( \dfrac{f(a+h) - f(a)}{h} \) as the difference quotient.
More will be said about the difference quotient in the coming days but for now, allow us to compute a few more difference quotients.
Compute the difference quotient of \( f(x) = x \).
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\( 1 \).
Compute the difference quotient of \( f(x) = x^3 - x \).
- Answer
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\( 3a^2 + 3ah + h^2 -1 \).
Compute the difference quotient of \( f(x) = \sqrt{x} \). Hint: to simplify, rationalize the numerator!
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\( \frac{1}{ \sqrt{a+h} - \sqrt{a}} \).
Compute the difference quotient of \( f(x) = \dfrac{1}{x} \).
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\( - \dfrac{1}{(a+h)(a)} \).
Compute the difference quotient of \( f(x) = \dfrac{1-x}{2+x} \).
- Answer
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\( \dfrac{-3}{(2+a+h)(2+a)} \).