# F'(x) existent or not.

#### Robokapp

It's a question I had on a quiz a few minutes ago.

f(x) = sin(x) for x < or = 0 and
f(x) = x for x > 0

Question was...does f'(x) exist? what is it?

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First I proved the f(x) is continuous at 0 by stating limit as x->0 from left = limit as x->0 from right = f(0).

Then...taking the derivative turned out to be a problem.
the lim[f(x+h)-f(x)]/h was unclear because...if x=0 and you add some h to it, that'd mean you use as your f(x) the x...but point (0, 0) belongs to Sin(x) not to x.

What i did is state that in x-values close to zero, from left and from right, the f'(x) = 1. Basically I ruled out situations like |x| where dy/dx = -1 and 1 for values around x=0.

I assumed that if f'(-0.0001) and f'(0.0001) are equal, the function (who we know is continuous) must be differentiable at that point.

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But my question is...is my logic something that will make my teacher grab his hairs before giving me a nice zero? or a 'wise' way to look the issue?

Also, is there a better, clearer way to solve this?

I always experience uncertainty on what to do and to which expression when a function gets split according to Domain.

Thank you.

~Robokapp

#### arildno

Homework Helper
Gold Member
Dearly Missed
Split your argument in two:
1. Find the right-hand derivative
2. Find the left-hand derivative

If they are equal, then your function is differentiable at 0.

#### matt grime

Homework Helper
f is trivially differentiable for x>0 and x<0. At x=0, one merely needs to verify if the left and right derivatives exist and agree. They trivially exist and as cos(x) tends to 1 as x tends to zero they agree.

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