If f(x)=2+abs(x-3) for all x, what's the value of the derivative f'(x) at x=3? (Answer: Nonexistent)

Question

solomonzelman · Answer

I've never a person with   (x-3)   abbs.

anonymous · Answer

abs means absolute value of something.

shamil98 · Answer

$$f(x) = 2 + |x-3|$$

anonymous · Answer

That's right. Can you continue?

shamil98 · Answer

Find the derivative, and then evaluate it at f'(3).
I assume you need assistance finding the derivative , correct?

anonymous · Answer

Yes. Please find the derivative.

anonymous · Answer

This will help you: http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/Calculus/DerivativesInvolvingAbsoluteValue.pdf

shamil98 · Answer

Okay, let u = x-3
|u|  = √u^2

$$u * \frac{ u' }{ |u| }$$

shamil98 · Answer

sub back in
(x-3)' = 1
u = x-3
|u| = |x-3|
$$\huge \frac{ x-3 }{ |x-3| }$$

shamil98 · Answer

I've never done derivatives involving absolute values, so I apologize in advance if this is incorrect, just my attempt.

shamil98 · Answer

From what I've gathered, the basic rule for this is given through:
The derivative of |x| is x/|x|. 

Applying the chain rule to this, the derivative of a function |u| is 
(u/|u|) * u'

anonymous · Answer

So (x-3)/(abs(x-3)) at x=3 is 0/0, which is 0, right?

shamil98 · Answer

Ah, yes I forgot about the actual question, evaluating it at f'(3).

shamil98 · Answer

0/0=undefined

shamil98 · Answer

It is indeterminate, (i think that is the term).

anonymous · Answer

So undefined means nonexistent, right?

shamil98 · Answer

Yes, it is non-existent, unless you were asked to solve for the indeterminacy .

anonymous · Answer

Thank you. You've helped a lot.