Questions: The Inverse Function Theorem

Question 1

Let F(r, θ) = (r cos θ, r sin θ). At which points does the inverse function theorem guarantee a local C¹ inverse?

• A. At every point
• B. At points with r ≠ 0
• C. At points with θ ≠ 0
• D. Only at the point (1, 0)

Answer: (B)

The Jacobian determinant of polar coordinates is r. It is nonzero exactly when r ≠ 0, so F has a local C¹ inverse near every point with r ≠ 0. At r = 0 every θ maps to the origin, so no local inverse exists there.

Question 2

If f: ℝ² → ℝ² is C¹ and det Df(x) ≠ 0 for every x, then f is one-to-one on ℝ².

• T. True
• F. False

Answer: False

The theorem is local. The map f(x, y) = (eˣ cos y, eˣ sin y) has det Df = e²ˣ, which is never zero, yet f(x, y) = f(x, y + 2π). Each point has a neighborhood on which f is one-to-one, but f is not one-to-one globally.

Question 3

Suppose f is C¹ near a, Df(a) is invertible, and b = f(a). Express D(f⁻¹)(b) in terms of Df.

Model answer: D(f⁻¹)(b) = [Df(a)]⁻¹

Differentiate f⁻¹(f(x)) = x with the chain rule at x = a: D(f⁻¹)(b) · Df(a) = I. So the derivative of the inverse is the inverse matrix of the derivative, the n-dimensional version of (f⁻¹)'(b) = 1/f'(a).

Question 4

The function f(x) = x³ shows that the hypothesis det Df(a) ≠ 0 is not necessary for f to be one-to-one near a. What conclusion of the theorem fails for x³ at a = 0?

Model answer: The inverse x^(1/3) is not differentiable at 0.

x³ is a bijection of ℝ, but its inverse has infinite slope at 0. The condition Df(a) invertible is exactly what the inverse needs in order to be differentiable at f(a).