322 Chapter 6 • Differentiable Functions
yxFigure 6.6Proof. Suppose f is differentiable on (a, b) and continuous on [a, b], where
a< b, and f(a) = f(b).
Case 1 (f is constant on [a, b]): In this case, we can take any c E (a, b); for then
f'(c) = 0, by Corollary 6.1.3.
Case 2 (f is not constant on [a, b]): By the extreme value theorem (5.3.7),
3 c,d E [a,b] 3
f(c) = minf[a,b] and f(d) = maxf[a,b],and since f is not constant on [a, b], either f(c) < f(a) = j(b) or f(d) > f(a) =
f(b).
Subcase 2a (f(c) < f(a) = f(b)): Then c ":fa a and c ":fa b. Thus c E (a, b)
and f(c) = min f[a, b] = min f(a, b). That is , f has a local minimum at c.
Therefore, by Theorem 6.3.4, f'(c) = 0.
Subcase 2b (f(d) > f(a) = f(b)): Exercise l. •
Example 6.4.2 Use Rolle's theorem to prove that the equation 5x^3 - 2x^2 +
x - 56 = 0 cannot have more than one real root. '
Solution. Let f(x) = 5x^3 - 2x^2 + x - 56. If the equation f(x) = 0 has
more than one real number solution, say x 1 and x 2 where x 1 < x 2 , then by
Rolle's theorem, :Jc E (x 1 ,x 2 ) 3 f'(c) = 0. But f'(x) = 15x^2 - 4x + l. The
discriminant of this quadratic is D = (-4)^2 -4(15)(1) < 0. Hence, there are no
real numbers x for which f'(x) = 0. Therefore, the equation f(x) = 0 cannot
have more than one real number solution. D
The following theorem is one of the most powerful in the entire calculus of
derivatives.