Lemma 4.4.1: If I, J ⊆ ℝ are intervals and f : ℝ → ℝ is strictly
monotone on I, maps I onto J, is differentiable at x₀ ∈ I, and
f' x₀ ≠ 0, then the inverse satisfies
(f ⁻¹)' (f x₀) = 1 / f' x₀. If moreover f is continuously differentiable
with derivative never zero on I, then the inverse is continuously
differentiable on J.
Lemma 4.4.1 (smoothness of the inverse): If f : ℝ → ℝ is continuously
C¹, strictly monotone on an interval I, maps I onto another interval J,
and derivWithin f I x ≠ 0 for every x ∈ I, then the inverse function is
continuously differentiable on J.
Theorem 4.4.2 (inverse function theorem). Given a continuously
C¹ function f : ℝ → ℝ on (a, b) with derivWithin f (Set.Ioo a b) x₀ ≠ 0
at some x₀ ∈ (a, b), there exists an open interval I ⊆ (a, b) containing
x₀ such that f restricts to an injective map on I with a continuously
differentiable inverse g : J → I on J = f '' I, and for every y ∈ J we have
derivWithin g J y = (derivWithin f (Set.Ioo a b) (g y))⁻¹.
Corollary 4.4.3: For n ∈ ℕ and x ≥ 0 there is a unique y ≥ 0
with y ^ n = x. The function g x = x^{1/n} on (0, ∞) is continuously
differentiable with derivative g' x = (1 / n) * x ^ ((1 - n) / n),
where fractional powers use the convention x ^ (m / n) = (x ^ (1 / n)) ^ m.
Example 4.4.4: For f x = x^2, the derivative at x₀ ≠ 0 is nonzero.
At a positive point x₀, the inverse function theorem applies on the maximal
open interval (0, ∞) where f is injective; any open set on which f is
injective and contains x₀ must lie in (0, ∞).
Example 4.4.5: The cubic f x = x^3 : ℝ → ℝ is bijective with a
continuous derivative, so it has a set-theoretic inverse g y = y^{1/3} on
all of ℝ. Nevertheless the inverse fails to be differentiable at 0
because f' 0 = 0, and the graph of the cube root has a vertical tangent at
the origin.