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8. DIFFERENTIATION OF REAL VALUED FUNCTIONS
8.2.12. Problem. Let $T_{a}:x\mapsto x+a$. The mapping $T_{a}$ is called translation by $a$. Note that it is {\it not} linear (unless, of course, $a=0$). We say that functions $f$ and $g$ in $\mathcal{F}_{a}$ are tangent at $a$ if the functions $f\mathrm{o}T_{a}$ and $g\mathrm{o}T_{a}$ are tangent at zero.
(a) Let $f(x)=3x^{2}+10x+13$ and $g(x)=\sqrt{-20x-15}$. Show that $f$ and $g$ are tangent at $-2.$
(b) Develop a theory for the relationship (`tangency at $a$'' which generalizes our work on ``tangency at $0$'' .
8.2.13. Problem. Each of the following is an abbreviated version of a proposition. Formulate precisely and prove.
(a) $C_{0}+\mathfrak{O}\subseteq C_{0}.$
(b) $C_{0}+0\subseteq C_{0}.$
(c) $\mathfrak{O}+0\subseteq \mathfrak{O}.$
8.2.14. Problem. Suppose that $f\simeq g$. Then the following hold.
(a) If $g$ is continuous at $0$, so is $f.$
(b) If $g$ belongs to $\mathfrak{O}$, so does $f.$
(c) If $g$ belongs to $0$, so does $f.$
8.3. LINEAR APPROXIMATION
One often hears that differentiation of a smooth function $f$ at a point $a$ in its domain is the process of finding the best ``linear approximation'' to $f$ at $a$. This informal assertion is not quite correct. For example, as we know from beginning calculus, the tangent line at $x=1$ to the curve $y=4+x^{2}$ is the line $y=2x+3$, which is not a linear function since it does not pass through the origin. To rectify this rather minor shortcoming we first translate the graph of the function $f$ so that the point $(a,\ f(a))$ goes to the origin, and {\it then} find the best linear approximation at the origin. The operation of translation is carried out by a somewhat notorious acquaintance from beginning calculus $\triangle y$. The source of its notoriety is two-fold: first, in many texts it is inadequately defined; and second, the notation $\triangle y$ fails to alert the reader to the fact that under consideration is a function of {\it two} variables. We will be careful on both counts.
8.3.1. Definition. Let $f\in \mathcal{F}_{a}$. Define the function $\triangle f_{a}$ by
$$
\triangle f_{a}(h):=f(a+h)-f(a)
$$
for all $h$ such that $a+h$ is in the domain of $f$. Notice that since $f$ is defined in a neighborhood of $a$, the function $\triangle f_{a}$ is defined in a neighborhood of $0$; that is, $\triangle f_{a}$ belongs to $\mathcal{F}_{0}$. Notice also that $\triangle f_{a}(0)=0.$
8.3.2. Problem. Let $f(x)=\cos x$ for $0\leq x\leq 2\pi.$
(a) Sketch the graph of the function $f.$
(b) Sketch the graph of the function $\triangle f_{\pi}.$
8.3.3. Proposition. {\it If} $f\in \mathcal{F}_{a}$ {\it and} $\alpha\in \mathbb{R}$, {\it then}
$$
\triangle(\alpha f)_{a}=\alpha\triangle f_{a}.
$$
Proof. To show that two functions are equal show that they agree at each point in their domain. Here
$$
\triangle(\alpha f)_{a}(h)=(\alpha f)(a+h)-(\alpha f)(a)
$$
$$
=\alpha f(a+h)-\alpha f(a)
$$
$$
=\alpha(f(a+h)-f(a))
$$
$$
=\alpha\triangle f_{a}(h)
$$
for every $h$ in the domain of $\triangle f_{a}.$
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