Introduction to Mathematical Analysis I - Second Edition

115 Figure 4.5: Strictly Increasing Function. Example 4.3.1 Let n ∈ N and f : [ 0 , ∞ ) → R be given by f ( x ) = x n . Then f 0 ( x ) = nx n − 1 . There- fore, f 0 ( x ) > 0 for all x > 0 and, so, f is strictly increasing. In particular, this shows that every positive real number has exactly one n -th root (refer to Example 3.4.2 ) . Theorem 4.3.3 — Inverse Function Theorem. Suppose f is differentiable on I = ( a , b ) and f 0 ( x ) 6 = 0 for all x ∈ ( a , b ) . Then f is one-to-one, f ( I ) is an open interval, and the inverse function f − 1 : f ( I ) → I is differentiable. Moreover, ( f − 1 ) 0 ( y ) = 1 f 0 ( x ) , (4.7) where f ( x ) = y . Proof: It follows from Theorem 4.2.5 that f 0 ( x ) > 0 for all x ∈ ( a , b ) , or f 0 ( x ) < 0 for all x ∈ ( a , b ) . Suppose f 0 ( x ) > 0 for all x ∈ ( a , b ) . Then f is strictly increasing on this interval and, hence, it is one-to-one. It follows from Theorem 3.4.10 and Remark 3.4.11 that f ( I ) is an open interval and f − 1 is continuous on f ( I ) . It remains to prove the differentiability of the inverse function f − 1 and the representation of its derivative ( 4.7 ) . Fix any ¯ y ∈ f ( I ) with ¯ y = f ( ¯ x ) . Let g = f − 1 . We will show that lim y → ¯ y g ( y ) − g ( ¯ y ) y − ¯ y = 1 f 0 ( ¯ x ) . Fix any sequence { y k } in f ( I ) that converges to ¯ y and y k 6 = ¯ y for every k . For each y k , there exists x k ∈ I such that f ( x k ) = y k . That is, g ( y k ) = x k for all k . It follows from the continuity of g that { x k } converges to ¯ x . Then lim k → ∞ g ( y k ) − g ( ¯ y ) y k − ¯ y = lim k → ∞ x k − ¯ x f ( x k ) − f ( ¯ x ) = lim k → ∞ 1 f ( x k ) − f ( ¯ x ) x k − ¯ x = 1 f 0 ( ¯ x ) .