103 ■ Example 4.3.1 Let n∈Nand f : [0,∞) →Rbe 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 f ′(x) , (4.7) 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 (4.7). Fix anyy0 ∈ f (I) withy0 =f (x0). Let g=f − 1. We will show that lim y→y0 g(y)−g(y0) y−y0 = 1 f ′(x0) . Fix any sequence {yn}in f (I) that converges toy0 and yn̸ =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 gthat {xn} converges to x0. Then lim n→∞ g(yn)−g(y0) yn −y0 =lim n→∞ xn −x0 f (xn)−f (x0) =lim n→∞ 1 f (xn)−f (x0) xn −x0 = 1 f ′(x0) . The proof is now complete. □ ■ Example 4.3.2 Let n∈Nand consider the function f : (0,∞) →Rgiven by f (x) =x n. Then f is differentiable and f ′(x) =nxn−1̸ =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 f ′(f −1(y)) = 1 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 f −1(y) the (positive) nth root of y and denote it by n √y. We also obtain the formula (f −1)′(y) = 1 n( n √y)n−1 .
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