Calculus: Analytic Geometry and Calculus, with Vectors

288 Functions, graphs, and numbers

5.27, which we do not bother to read now, proves that this idea is correct

and that a < b. We next mark xo such that a < xo < b. We can

easily believe that the rising graph of x(t) must intersect the dotted

horizontal line at exactly one point (to,xo). Theorems 5.48 and 5.27 prove

that this is correct, and we now have to. Our given function y, of which

we have not sketched a schematic graph, then determines the number yo

defined by yo = y(to), and we put yo = f(xo) in

Figure 5.173. This gives one point on the graph

of y = f(x), and the same procedure gives each

other point on the graph. We now have the

formula y(t) = f(x(t)). If we had proof that f

is differentiable, we could apply the chain rule

o a xo b x to obtain y'(t) = f'(x(t))x'(t) and divide by x'(t)

Figure 5.173 to get our answer, but this will not work because

we do not yet have the required proof. We

therefore start a direct attack upon difference quotients by taking a

fixed t for which tl < t < t2 and writing

(5.174) f(x(t + At)) - f(x(t)) = Y(t + At) - Y(t)-

Dividing by x(t + At) - x(t) gives the more promising formula

(5.175) f(x(t + At)) - f(x(t))=Y(t + At) - Y(t)

x(t + At) - x(t) x(t + At) - x(t)

Since y'(t) and x'(t) both exist and x'(t) 0, we can divide the numerator

and denominator of the right side by At and see that the right side has the

limit y'(t)/x'(t) as At--+ 0. The left side therefore has the same limit

and we obtain

(5.176) lim

f (x(t + At)) - f (x(t)) y'(t)

Aim [ x(t + At) - x(t) j

_

x(t)

This seems to be almost the desired result (5.171), but we must use it to

obtain additional information. Let e > 0. Choose a positive number

Sl such that

(5.177) f(x(t + At)) - f(x(t))_Y'(t)

x(t + At) - x(t) x'(t) < e

whenever jAtj < a,. Another appeal to theorems given later in this

chapter shows that there is a positive number S2 such that when Jkl < S2,

there is a number At for which jAti < Si and

(5.178)

It follows that

(5.179)

x(t + At) = x(t) + A.

f(x(t) + h) - f((x(t)) -y'(t)

h XI(t) <e