Exercise 5.2.9

Decide whether each conjecture is true or false. Provide an argument for those that are true and a counterexample for each one that is false.

(a)
If f ′ exists on an interval and is not constant, then f ′ must take on some irrational values.
(b)
If f ′ exists on an open interval and there is some point c where f ′ ( c ) > 0 , then there exists a δ -neighborhood V δ ( c ) around c in which f ′ ( x ) > 0 for all x ∈ V δ ( c ) .
(c)
If f is differentiable on an interval containing zero and if lim ⁡ x → 0 f ′ ( x ) = L , then it must be that L = f ′ ( 0 ) .

Answers

(a)
If f ′ is not constant there exist x , y with f ′ ( x ) < f ′ ( y ) , since derivatives obey the intermediate value proper (Theorem 5.2.7) f ′ takes on the value of every irrational number in ( f ′ ( x ) , f ′ ( y ) ) .
(b)
True if f ′ is continuous, False in general. Let f ( x ) = x 2 sin ⁡ ( 1 ∕ x ) + x ∕ 2 so that f ′ ( x ) = 2 x sin ⁡ ( 1 ∕ x ) − cos ⁡ ( 1 ∕ x ) + 1 ∕ 2

Notice how f ′ alternates between positive and negative for small x . We have

lim ⁡ x → 0 x 2 sin ⁡ ( 1 ∕ x ) − 0 x − 0 = lim ⁡ x → 0 x sin ⁡ ( 1 ∕ x ) = 0

Thus f ′ ( 0 ) = 1 ∕ 2 , pick any δ and I can find x ∈ V δ ( 0 ) with f ′ ( x ) ≤ 0 . To be explicit define x n = 1 ∕ ( 2 πn ) so that f ′ ( x n ) = − 1 ∕ 2 then pick n large enough so x n ∈ V δ ( 0 ) .

(c)
A direct proof using L’Hospital’s rule: continuity of f forces the limit of the quotient to be 0 ∕ 0 . f ′ ( 0 ) = lim ⁡ x → 0 f ( x ) − f ( 0 ) x = lim ⁡ x → 0 f ′ ( x ) = L

In general a corollary of L’Hospital’s rule is that derivatives can only have essential discontinuities, discontinuities where lim ⁡ x → c f ′ ( x ) doesn’t exist.

Alternatively you can see this by contradiction. Derivatives obeying the IVP rules out jump discontinuities, and removable discontinuities may be ruled out by 𝜖 - δ ing.

User profile picture
2022-01-27 00:00
Comments