Calculus: Analytic Geometry and Calculus, with Vectors

5.5 The Rolle theorem and the mean-value theorem 325

Figure 5.55

joining the points (a, f(a)) and (b, f(b)) of Figure 5.55. The point-slope
form of the equation of a line gives the formula

(5.561) g(x) - f(a) =f(b b-f(a)(x- a)

a

in which g(x) appears instead of the more familiar y. Hence,

(5.562) 0(x) = f(x) - f(a) - f(bb)b

af

(a) (x - a)

when a<x5band

(5.563) 4'(x) = f' (x)-f(bb

• a(a)

when a < x < b. It is easily seen that 0 satisfies the hypotheses of the

Rolle theorem. Choosing x* such that a < x* < b and q5'(x*) = 0 gives

the required conclusion (5.53). Multiplying by (b - a) then gives

(5.54), and Theorem 5.52 is proved.

Theorem 5.57 If f is continuous over a < x <_ b and f(x) = 0 when

a <x <b,then f(x) = f(a) when a <x <b.

To prove this theorem, we note first that f(x) = f(a) when x = a. If

a < xi < b, we can apply the mean-value theorem to the interval

a < x S xi to conclude that there is a number x* for which a < x* < xi

and

f (xi) - f (a) = f (x*) (xi - a).

But f'(x*) = 0 and hence f(xi) = f(a). Therefore, f(x) = f(a) when
a < x 5 b and Theorem 5.57 is proved. It follows from this theorem that
if two functions Fl and F2 have the same derivative over an interval, say
Fi(x) = F2(x) = g(x) when a < x < b, and we put f(x) = F2(x) - Fi(x),
then f'(x) = 0 when a < x < b so f(x) must be a constant c and

Fs(x) = Fi(x) + c

when a 5 x < b. This proves Theorem 4.13.