Principles of Mathematics in Operations Research

12.2 Mean Value Theorems 171

Theorem 12.2.3 Suppose f : [a,b] H->- R is differentiable and f'(a) < A <

f'(b) [/'(a) > A > /'(&)]. Then, 3x e (a, 6) 3 /'(a;) = A.

Proof. Let g(<) = /(£) - Xt. Then, #'(a) < 0 [g'(a) > 0] so that g(h) <

g(a) [g(ti) > g(a)] for some *i e (a,6), so that g(f 2 ) < g{b) [g{h) > #(a)l

for some £ 2 G (a, b). Hence, g attains its minimum [maximum] on [a,b] at

some points x 6 (a,b). By the first mean value theorem, g'(x) = 0. Hence,

f'{x) = X. D

Corollary 12.2.4 // / is differentiable on [a,b], then f cannot have any

simple discontinuities on [a,b].

Remark 12.2.5 But f may have discontinuities of the second kind.

Theorem 12.2.6 (L'Hospital's Rule) Suppose f and g are real and dif-

ferentiable in (a,b) and g'(x) ^ 0, Vx € (a,b) where oo < a < b < +oo.

Suppose

,;. -> A as x -> a (o).

ff'(x)

// /(x) —> 0 and g(x) —> 0 as x —> a or i/ /(x) -> +oo and g(x) —> +oo as

x —> a, i/ien

#(»)

4 OS I -> O.

Proof. Let us consider the case —oo < ^4 < +oo: Choose g 6 R 3 A < g, and

choose r 9 A < r < q. By (o),

fix)

3c E (a, b) 3 a < x < c => :L-y-~ < r. (£)

g(x)

If a < x < y < c, then by the second mean value theorem,

*6(M),&ia.m<r.(»,

g(x)-g{y) g'(y)

Suppose /(x) -> 0 and g(x) -» 0 as x -> a. Then, (4) |M <r<q,a<y<c.

Suppose g(x) -* +oo as x —> a. Keeping y fixed, we can choose c\ £ (a,y) 3

g{x) > g{y) and g(x) > 0 if a < x < cx. Multiplying (4) by \g(x)-g(y)]/g(x),
we have $g < r - rf(g + $g, o < z < cx. If a: -• a 3c 2 € (a,Cl) 9 $g <
q, a < x < C2. Summing with (4») Vo 3 A < q yields

f(x\

3c 2 3 -j-r < q if a < x < c 2.

Similarly, if —oo < A < +oo and p 3 p < A, 3c^ 3 p < :4fy, a < x < C3. D