NETMaths
Part CCSIR NET June 2023implicit-function-theorem-fails

Implicit function theorem fails

Consider the function defined by . Which of the following statements are true?

  1. A.There is no continuous real-valued function g defined on any interval of containing 0 such that f(x, g(x)) = 0.
  2. B.There is exactly one continuous real-valued function g defined on an interval of containing 0 such that f(x, g(x)) = 0.
  3. C.There is exactly one differentiable real-valued function g defined on an interval of containing 0 such that f(x, g(x)) = 0.
  4. D.There are two distinct differentiable real-valued functions g on an interval of containing 0 such that f(x, g(x)) = 0.

Solution

forces g(x) = (the real cube root is a bijection), so there is exactly one continuous solution. It is not differentiable at 0, so no differentiable g exists. The implicit function theorem does not apply since f_y(0,0) = 0.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Related counterexample: A C¹ map with everywhere non-zero Jacobian is injective

From Functions of Several VariablesInverse and implicit function theorems, extrema

ShareWhatsAppTelegram