What's wrong with these functions?

Calculus Level 3

We know that if x x is a real number than ∣ x ∣ 2 = x 2 |x|^2=x^2 . Now let's define f : R → R , f ( x ) = x 2 g : R → R , g ( x ) = ∣ x ∣ 2 . f:\mathbb{R}\rightarrow \mathbb{R}, \quad f(x)=x^2 \\ g:\mathbb{R}\rightarrow \mathbb{R}, \quad g(x)=|x|^2.

That being said, where have I made a mistake?

  • I f ( x ) = g ( x ) , ∀ x ∈ R f(x)=g(x), \, \forall x \in\mathbb{R} so f f and g g are exactly the same function (note that they share the same domain)
  • II f = g ⇒ f ′ = g ′ f=g \Rightarrow f'=g'
  • III f ′ f' and g ′ g' must have the same domain
  • IV f ′ ( x ) = 2 x , ∀ x ∈ R f'(x)=2x, \, \forall x \in\mathbb{R} while g ′ ( x ) = 2 ∣ x ∣ x ∣ x ∣ = 2 x , ∀ x ∈ R ∖ { 0 } g'(x)=2|x|\frac{x}{|x|}=2x, \, \forall x\in\mathbb{R} \smallsetminus \{0\}
Passage I Passages I & III Passage IV Passage III Passage II

Vittorio Bertoletti by Vittorio Bertoletti , 26, Italy

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1 solution

In Passage IV the chain rule has been applied to find g ′ g' but that was NOT legit. Sure g g can be seen as the composition of two functions but, in this case, the chain rule does not hold because the inner function (the absolute value function) is such that its derivative is not defined for x = 0 x=0 .

g ′ ( x ) = 2 x , ∀ x ∈ R g'(x)=2x, \, \forall x\in\mathbb{R} because we have to apply the definition of derivative instead of the chain rule.

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