103 ■ Example 4.3.1 Let n ∈ N and f : [0,∞) → R be given by f (x)= x n . Then f ′ (x)= nxn−1. Therefore, f ′ (x) > 0 for all x > 0 and, so, f is strictly increasing. In particular, this shows that every positive real number has exactly one nth root (refer to Example 3.4.2). Theorem 4.3.3 — Inverse Function Theorem. Suppose f is differentiable on I = (a,b) and f ′ (x)≠ 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) ′ (y)= 1 , (4.6) f ′ (x) where f (x)= y. Proof: It follows from Theorem 4.2.5 that f ′ (x) > 0 for all x ∈ (a,b), or f ′ (x) < 0 for all x ∈ (a,b). Suppose f ′ (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 3.4.7 and Remark 3.4.2 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 (4.6). Fix any y0 ∈ f (I) with y0 = f (x0). Let g = f − 1. We will show that g(y) − g(y0) 1 lim = . y→y0 y − y0 f ′ (x0) Fix any sequence {yn} in f (I) that converges to y0 and yn ̸ = y0 for every n. For each yn, there exists xn ∈ I such that f (xn)= yn. That is, g(yn)= xn for all n. It follows from the continuity of g that {xn} converges to x0. Then g(yn) − g(y0) xn − x0 1 1 lim = lim = lim = . n→∞ yn − y0 n→∞ f (xn) − f (x0) n→∞ f (xn) − f (x0) f ′ (x0) xn − x0 The proof is now complete. □ ■ Example 4.3.2 Let n ∈ N and consider the function f : (0,∞) → R given by f (x)= x n . Then f is differentiable and f ′ (x)= nxn−1 ≠ 0 for all x ∈ (0,∞). It is also clear that f ((0,∞)) = (0,∞). It follows from the inverse function theorem that f −1 : (0,∞) → (0,∞) is differentiable and given y ∈ (0,∞) ( f −1) ′ (y)= 1 = 1 f ′ ( f −1(y)) n( f −1(y))n−1 . Given y > 0, the value f −1(y) is the unique positive real number whose nth power is y. We call n f −1(y) the (positive) nth root of y and denote it by √ y. We also obtain the formula ( f −1) ′ (y)= n( √ n 1 y)n−1 .
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