Question

Question:
A minimum with no derivative :
The function f(x)=|x| has an absolute minimum value at x=0 even though f is not differentiable at x=0. Is this consistent with Theorem 2? Give reasons for your answer.

can someone explain briefly.. thank you..

Answer 1

THEOREM 2 : The First Derivative Theorem for Local Extreme Values
if f has a local maximum or minimum value at an interior point c of its domain, and if f' is defined at c, then ... f'(c)=0


Note:
Be careful not to misinterpret Theorem 2 because its converse is false. A differentiable function may have a critical point at x=c without having a local extreme value there. For instance, the function f(x)=x^3 has a critical point at the origin and zero value there, but is positive to the right of the origin and negative to the left. So it cannot have a local extreme value at the origin. Instead, it has a point of inflection here.

Answer 2

It is consistent with Theorem 2.
Look at the conditions to be met:
1. f has to have a local extreme in interrior point c of its domain.
OK: f(0) = 0 is a local minimum.
2. f' must be defined at c.
NOT OK: f'(0) is not defined because the left-derivative is not equal to the right derivative in x=0:\[\lim_{ x \downarrow 0}\frac{ f(x+h)-f(x) }{ h }=\lim_{ x \downarrow 0}\frac{ |h| }{ h }=\lim_{ x \downarrow 0}\frac{ h }{ h }=1\]and
\[\lim_{ x \uparrow 0}\frac{ f(x+h)-f(x) }{ h }=\lim_{ x \uparrow 0}\frac{ |h| }{ h }=\lim_{ x \uparrow 0}\frac{ -h }{ h }=-1\]So there is no such thing as THE limit of the differential quotient in x = 0, so there is no f'(0).

In short: because the conditions of Theorem 2 are not both met, the conclusion f'(c)=0 can't be drawn.

Answer 3

Thank you very much sir @ZeHanz !! , it comes clear...
this is a very good explaination.. i will jot it down... :)

Answer 4

Glad to be of help!