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8. DIFFERENTIATION OF REAL VALUED FUNCTIONS

8.2.12. Problem. Let Ta:xx+a. The mapping Ta is called translation by a. Note that it is not linear (unless, of course, a=0). We say that functions f and g in Fa are tangent at a if the functions foTa and goTa are tangent at zero.

(a) Let f(x)=3x2+10x+13 and g(x)=-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) C0+OC0.

(b) C0+0C0.

(c) O+0O.

8.2.14. Problem. Suppose that fg. Then the following hold.

(a) If g is continuous at 0, so is f.

(b) If g belongs to 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+x2 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 then find the best linear approximation at the origin. The operation of translation is carried out by a somewhat notorious acquaintance from beginning calculus y. The source of its notoriety is two‐fold: first, in many texts it is inadequately defined; and second, the notation y fails to alert the reader to the fact that under consideration is a function of two variables. We will be careful on both counts.

8.3.1. Definition. Let fFa. Define the function fa by

fa(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 fa is defined in a neighborhood of 0; that is, fa belongs to F0. Notice also that fa(0)=0.

8.3.2. Problem. Let f(x)=cosx for 0x2π.

(a) Sketch the graph of the function f.

(b) Sketch the graph of the function fπ.

8.3.3. Proposition. If fFa and αR, then

(αf)a=αfa.

Proof. To show that two functions are equal show that they agree at each point in their domain. Here

(αf)a(h)=(αf)(a+h)-(αf)(a)

=αf(a+h)-αf(a)

=α(f(a+h)-f(a))

=αfa(h)

for every h in the domain of fa.

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