Answer on Calculus Question for Luke
f (x) = cos x, x < 0
1 x ≥ 0.
Is f ′ continuous at 0?
When x tends to zero: x->0 f'(x) = -sin 0 = 0.
For x ≥ 0:
f'(x) = (1)' = 0
As f'(x) at x =0 equals to 0 to the both sides from the point, it's continuous.
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