# 7.2. Difference Quotient Functions

Now we are going to develop a general idea from the results of our introductory example. As the secant gradients played a major role there our first step will be to asign to any function  f, with respect to a preselected point a, a survey of all these secant gradients.

Definition:  Let $a\in A\subset ℝ$ and  $f:A\to ℝ$ be an arbitrary function. The function

 ${m}_{\phantom{\rule{0.0em}{0ex}}a}≔\frac{f-f\left(a\right)}{X-a}$ [7.2.1]

is called the difference quotient function of  f in respect of a.

Consider:

• ${m}_{\phantom{\rule{0.0em}{0ex}}a}:A\\left\{a\right\}\to ℝ$  and  ${m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .
As constructed, the preselected number a  is no member of the domain of any difference quotient function.

• Occasionally we use ${m}_{\phantom{\rule{0.0em}{0ex}}f,a}$ instead of  ${m}_{\phantom{\rule{0.0em}{0ex}}a}$ to focus on the relation to  f.

• The values of  ${m}_{\phantom{\rule{0.0em}{0ex}}a}$ are the gradient numbers of the secants.

Sometimes the value ${m}_{a}\left(x\right)$ is used for measuring the alteration behaviour of a function and is called the rate of change (or average rate of change) of  f between a and x in this context.

• A special notation is used in physics: When studying a moving particle for example, the distance covered within t units of time is usually denoted by $s\left(t\right)$. The rates of change $\frac{s\left({t}_{2}\right)-s\left({t}_{1}\right)}{{t}_{2}-{t}_{1}}$ respective to the function $t↦s\left(t\right)$, i.e. the quotients

$\frac{\text{distance covered}}{\text{time spent}}$,

are usually denoted by $\mathrm{\Delta }\phantom{\rule{0.1em}{0ex}}v=\frac{\mathrm{\Delta }\phantom{\rule{0.1em}{0ex}}s}{\mathrm{\Delta }\phantom{\rule{0.1em}{0ex}}t}$  i v for velocity from the Latin word velocitas.
and are addressed as the particle's average speed between the timepoints ${t}_{1}$ and ${t}_{2}$ or the waypoints $s\left({t}_{1}\right)$ and $s\left({t}_{2}\right)$ respectively.

The following example shows the difference quotient functions of several standard maps. The transformations carried out in addition are used in the next chapter.

Example:  We calculate the difference quotient functions in respect of a
• for a linear function  $mX+b$  with an arbitrary a:
 ${m}_{\phantom{\rule{0.0em}{0ex}}a}=\frac{mX+b-\left(ma+b\right)}{X-a}=\frac{m\left(X-a\right)}{X-a}=m\frac{X-a}{X-a}$ . [7.2.2]

• for a monomial  ${X}^{n}$  with arbitrary a and $n\in ℕ$ :
 [7.2.3]

We show the equality (*) by induction on n.

• for the reciprocal function  $\frac{1}{X}$ with $a\ne 0$ :
 ${m}_{\phantom{\rule{0.0em}{0ex}}a}=\frac{\frac{1}{X}-\frac{1}{a}}{X-a}=\frac{a-X}{aX\left(X-a\right)}=-\frac{X-a}{aX\left(X-a\right)}$ . [7.2.4]

• for the root function  $\sqrt{X}$ with $a\ge 0$ :
 ${m}_{\phantom{\rule{0.0em}{0ex}}a}=\frac{\sqrt{X}-\sqrt{a}}{X-a}=\frac{X-a}{\left(\sqrt{X}+\sqrt{a}\right)\left(X-a\right)}$ . [7.2.5]

• for the absolute value function  $|X|$  for $a=0$ :
 ${m}_{\phantom{\rule{0.0em}{0ex}}0}=\frac{|X|-|0|}{X-0}=\frac{|X|}{X}$ . [7.2.6]

It is always rewarding to check how far newly built notions cooperate with basic arithmetics. In many cases this leads to powerful technics. The following proposition shows that the mapping  $f↦{m}_{\phantom{\rule{0.0em}{0ex}}f,a}$  respects all of them.

Proposition:  Let  $A,B\subset ℝ$ and $a\in A\cap B$. For  $f:A\to ℝ$ and $g:B\to ℝ$ we have:

 1.${m}_{\phantom{\rule{0.0em}{0ex}}f+g,a}={m}_{\phantom{\rule{0.0em}{0ex}}f,a}+{m}_{\phantom{\rule{0.0em}{0ex}}g,a}$ [7.2.7]
 2.${m}_{\phantom{\rule{0.0em}{0ex}}f-g,a}={m}_{\phantom{\rule{0.0em}{0ex}}f,a}-{m}_{\phantom{\rule{0.0em}{0ex}}g,a}$ [7.2.8]
 3.${m}_{\phantom{\rule{0.0em}{0ex}}f\cdot g,a}={m}_{\phantom{\rule{0.0em}{0ex}}f,a}\cdot g+f\left(a\right)\cdot {m}_{\phantom{\rule{0.0em}{0ex}}g,a}$ [7.2.9]
 4.${m}_{\frac{f}{g},a}=\frac{{m}_{\phantom{\rule{0.0em}{0ex}}f,a}\cdot g-f\cdot {m}_{\phantom{\rule{0.0em}{0ex}}g,a}}{g\left(a\right)\cdot g}$ [7.2.10]

Proof:
1. ►  ${m}_{\phantom{\rule{0.0em}{0ex}}f+g,a}=\frac{\left(f+g\right)-\left(f+g\right)\left(a\right)}{X-a}=\frac{f-f\left(a\right)}{X-a}+\frac{g-g\left(a\right)}{X-a}={m}_{\phantom{\rule{0.0em}{0ex}}f,a}+{m}_{\phantom{\rule{0.0em}{0ex}}g,a}$ .

2. ►  The calculation is a copy of 1.

3. ►  We need the basic adding zero trick, this time: $0=-f\left(a\right)g+f\left(a\right)g$.

4. ►  Again we add zero now in the shape of  $0=fg-fg$.

 7.1. 7.3.