Calculus: Analytic Geometry and Calculus, with Vectors

162 Functions, limits, derivatives

15 A long time ago it was discovered that if P is a polynomial, then between
each pair of values of x for which P(x) = 0 there must be at least one value ofx
for which P'(x) = 0. For the special case in which

P(x) = (x - 1)(x - 2)(x - 3),

find the values of x for which P(x) = 0 and the values of x for which P'(x) = 0,

and verify the statement about the zeros of P and P. Remark: This matter will

become quite unmysterious when we learn about the Rolle theorem.

16 In connection with the definition of the derivative of a function f at a

point x, we recognized the possibility that this derivative may fail to exist.

To clarify this matter, we should know about the simplest example of a con-

tinuous function f which is not everywhere differentiable. To investigate such

matters, we should know about the right-hand derivative f+(x) and the left-hand

derivative f-'(x) that are defined by the formulas

f+(x) = lim

f(x + Ax) - f(x) f,(z) = lim f(x + Ax) - f(x)

er-.o+ Ax AZ-->o- Ax

when these limits exist. It is easy to guess and almost as easy to prove that f is

differentiable at x if and only if

f+(x) = f_ (x) = f'(x).

For the simplest example in whichf(x) = Jxl, show that ff(0) = 1 and f-'(0) _ -1

and hence that f'(0) does not exist. It is not so easy to construct a continuous

function which is everywhere nondifferentiable, but Weierstrass (1815-1897)

started this construction business a long time ago.

17 Construct and look at a graph similar to those in Figures 3.46 and 3.47

but having the loops tangent to the lines having equations y = x and y = -x.

Letting this be the graph off and, letting f(0) = 0, discuss continuity of f at 0

and discuss f' (0), f'_(0), and f'(0).

18 For reasons that may be partially explained by the remark at the end of

this set of problems, we give, in terms of the notation of Newton, a complete

statement and proof of the part of Theorem 3.56 that involves the product

formula (3.565).

(1) Theorem If g and h are functions differentiable at x and if f is the func-

tion for which

(2) f(x) = g(x)h(x)

when x belongs to the domains of g and h, then f is differentiable at x and

(3) f'(x) = g(x)h'(x) + h(x)B (x)

Proof: Since g and h are differentiable at x, there must be an interval I with

center at x over which g and h are defined. When x + Ax lies in this interval,

we have

(4) f(x + ,Ax) = g(x + Ax)h(x + Ax)