Mean Value Theorem

Theorem 3.2.1

If f is differentiable on an open interval (a,b) and f assumes its maximum or minimum at a point c \in (a,b), then f^\prime(c) = 0.

Suppose that f assumes its maximum at c. That is, f(x) \leq f(c) for all x \in (a,b). Let (x_n) be a sequence converging to c such that a < x_n < c for all n. Since f is differentiable at c, the sequence

\left( \frac{f(x_n) - f(c)}{x_n - c} \right)

converges to f^\prime(c). Since f(x_n) \leq f(c) and x_n \leq c, each term in our above sequence is nonnegative. Thus f^\prime(c) \geq 0.

We repeat this argument with a sequence (y_n), such that c < y_n < b for all n. Each term in the sequence

\left( \frac{f(y_n) - f(c)}{y_n - c} \right)

will be nonpositive, so f^\prime(c) \leq 0. We therefore conclude that f^\prime(c) = 0. If f has a minimum at c, we apply the baove results to the function -f.

Theorem 3.2.2 (Rolle’s Theorem)

Let f be a continuous function on [a,b] that is differentiable on (a,b) and such that f(a) = f(b). Then there exists at least one point c in (a,b) such that f^\prime(c) = 0.

Since f is continuous on a compact interval [a,b], f obtains a minumum x_1 and maximum x_2 by the Extreme Value Theorem.

If x_1,x_2 are both endpoints of [a,b], then the function is a constant, and f^\prime(x) = 0 for all x \in [a,b].

Otherwise, f assumes either a minimum or maximum at some point c \in (a,b), and by Theorem 3.2.1, f^\prime(c) = 0.

Theorem 3.2.3 (Mean Value Theorem)

Let f be a continuous function on [a,b] that is differentiable on (a,b). Then there exists at least one point c \in (a,b) such that

f^\prime(c) = \frac{f(b) - f(a)}{b-a}.

Let g(x) be a function whose graph is the chord between f(a) and f(b). More formally

g(x) = \frac{f(b)-f(a)}{b-a}(x-a) + f(a), \quad \text{for all $x \in [a,b]$}.

Then the function h = f - g is continuous on [a,b] and differentiable on (a,b). Since f(a) = g(a) and f(b) = g(b), we have that h(a) = h(b) = 0. Applying Rolle’s Theorem we see that for some c \in (a,b),

0 = h^\prime(c) = f^\prime(c) - g^\prime(c) = f^\prime(c) - \frac{f(b) - f(a)}{b-a}.

Thus

f^\prime(c) = \frac{f(b)-f(a)}{b-a}.

Theorem 3.2.4

Let f be a continuous function on [a,b] that is differentiable on (a,b). If f^\prime(x) = 0 for all x \in (a,b), then f is constant on [a,b].

Proof by Contradiction

Suppose f were not constant on [a,b]. Then there exists two points a \leq x_1 < x_2 \leq b such that f(x_1) \not = f(x_2). But then by MVT, for some c \in (x_1,x_2) there exists

f^\prime(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \not = 0

Thus a contradiction.

Corollary 3.2.5

Let f,g be continuous on [a,b] and differentiable on (a,b). Suppose that f^\prime(x) = g^\prime(x) for all x \in (a,b). Then there exists a constant C such that

f(x) = g(x) + C \quad \text{ for all $x \in [a,b]$}

Direct Proof

Apply Theorem 3.2.4 to -g.

Theorem 3.2.6

Let f be differentiable on I. Then

(a) if f^\prime(x) > 0 for all x \in I, then f is strictly increasing on I, and

(b) if f^\prime(x) < 0 for all x \in I, then f is strictly decreasing on I.

Direct Proof

Theorem 3.2.7 (Intermediate Value Theorem for Derivatives)

Let f be differentiable on [a,b] and suppose there exists a number k between f^\prime(a) and f^\prime(b). Then there exists a point c \in (a,b) such that f^\prime(c) = k.

(a) if f^\prime(x) > 0 for all x \in I, then f is strictly increasing on I, and

(b) if f^\prime(x) < 0 for all x \in I, then f is strictly decreasing on I.

Direct Proof