Calculus: Analytic Geometry and Calculus, with Vectors

584 Partial derivatives

lar region R of Figure 11.391 is a subset of the given rectangular region RI and,
moreover,

(14) G(x,yi) < 0 (xo - h < x < xo + h)

(15) G(x,y2) > 0 (xo - h < x < xo + h).

The negative and positive signs in Figure 11.391 serve to remind us that G(x,y) <

0 at points (x,y) on the lower edge of R and that G(x,y) > 0 at points (x,y) on

the upper edge of R. We are now pre-

(xo-h,y2) (X0' Y2) (xo+h,yz) pared to obtain results. Let xo - h <

x < xo + h. Since G0(x,y) > 0 when

y, S y 5 y2, we conclude that G(x,y) is

(x0 -h,Y1)

Figure 11.391

(xo, Yi) (xo+h,y,)

increasing over the interval y, S y S Y.

Since G(x,yi) < 0 and G(x,y2) > 0, we

conclude, with the aid of the intermedi-

ate-value theorem, that there is one

and only one number f (x) for which

yi < f(x) < y2 and G(x, f(x)) = 0.

Our proof is now about half done; we

have found our f but we need information about f'(x). To start getting this

information, suppose that xo - h < x < xo + h and xo - h < x + Ax < xo + h.

Then

(16) G(x, f(x)) = 0, G(x + Ax, f(x + Ax)) = 0

and consequently

(17) [G (x + Ax, f(x + Ax)) - G(x, f(x + Ax))]

+ [G(x, f(x + Ax)) - G(x, f(x))] = 0.

Applying the mean-value theorem (Theorem 5.52) to these differences shows that

there exists a number between x and x + Ax and a number 71 between f(x)

and f (x + Ax) such that

(18) G:(, f (x + Ax)) Ax +G.,(x,r7)[f(x + Ax) - Ax)] = 0

and hence

(19)

f(x+Ax) -f(x)__G,(,f(x+Ax))

Ax G,, (x,,?)

Since G. and G are, by hypothesis, continuous at the point (x, f(x)), the desired

result (11) will follow from this if we show that f is continuous at x, that is,

(20) lim f(x + Ax) = f(x).

To prove (20), let a be a positive number for which f (x) + e and f (x) - e lie

between yl and y2. Since G(x, f(x)) = 0 and G,,(x,y) is an increasing function

of y, we find that

G(x,f(x) - e) <0 <G(x,f(x)+e).

Since G is continuous, we can choose a positive number 8 such that

G(x + Ax, f (X) - e) < 0 < G(x -1- Ax, f (X) + e)