Introduction to Mathematical Analysis I - Second Edition

108 4.2 THE MEAN VALUE THEOREM Figure 4.1: Illustration of Fermat’s Rule. Proof: Suppose f has a local minimum at a . Then there exists δ > 0 sufficiently small such that f ( x ) ≥ f ( a ) for all x ∈ B ( a ; δ ) . Since B + ( a ; δ ) is a subset of B ( a ; δ ) , we have f ( x ) − f ( a ) x − a ≥ 0 for all x ∈ B + ( a ; δ ) . Taking into account the differentiability of f at a yields f 0 ( a ) = lim x → a f ( x ) − f ( a ) x − a = lim x → a + f ( x ) − f ( a ) x − a ≥ 0 . Similarly, f ( x ) − f ( a ) x − a ≤ 0 for all x ∈ B − ( a ; δ ) . It follows that f 0 ( a ) = lim x → a f ( x ) − f ( a ) x − a = lim x → a − f ( x ) − f ( a ) x − a ≤ 0 . Therefore, f 0 ( a ) = 0. The proof is similar for the case where f has a local maximum at a . Theorem 4.2.2 — Rolle’s Theorem. Let a , b ∈ R with a < b and f : [ a , b ] → R . Suppose f is continuous on [ a , b ] and differentiable on ( a , b ) with f ( a ) = f ( b ) . Then there exists c ∈ ( a , b ) such that f 0 ( c ) = 0 . (4.3) Proof: Since f is continuous on the compact set [ a , b ] , by the extreme value theorem (Theorem 3.4.2 ) there exist ¯ x 1 ∈ [ a , b ] and ¯ x 2 ∈ [ a , b ] such that f ( ¯ x 1 ) = min { f ( x ) : x ∈ [ a , b ] } and f ( ¯ x 2 ) = max { f ( x ) : x ∈ [ a , b ] } . Then f ( ¯ x 1 ) ≤ f ( x ) ≤ f ( ¯ x 2 ) for all x ∈ [ a , b ] . (4.4)