Derivatives of absolute values

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The function f(x) = |x| is differentiable for all x except at x = 0, where it has a cusp. For x > 0, f(x) behaves as f(x) = x, and for x < 0, it behaves as f(x) = -x, both of which are differentiable. The derivative is determined using the limit definition, which shows that at x = 0, the left-hand and right-hand limits do not match, indicating non-differentiability. Understanding the concept of a cusp is crucial for grasping why the function fails to be differentiable at this point. Overall, the function's piecewise nature highlights its differentiability except at the origin.
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Homework Statement


Where is the function f(x) = |x| differentiable?


Homework Equations


[f(x+h) - f(x)] / h


The Attempt at a Solution


I know that the graph of f(x)=|x| shows a corner at the origin from which 2 lines project at opposite slopes, as in they are symmetric about the y-axis.
I've seen the solution but I don't understand why f(x) is differentiable for every number except 0.
 
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|x|=x for x>0 and |x|=(-x) for x<0. x and -x are both differentiable. Remember the derivative expression is a limit. It's only correct in the limit where h->0. Think of h as REALLY SMALL.
 
You can re-write |x| as a piecewise function

f(x) = -x, x<0
f(x) = 0, x=0
f(x) = x, x>0

Find the derivative and I think you'll see your answer -- Oh and you might want to read up on what a "cusp" is.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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