Question #749

Using the deﬁnition of the derivative, determine f ′ (0) for
f (x) = cos x, x < 0
1 x ≥ 0.
Is f ′ continuous at 0?

Expert's answer

For x < 0 :f'(x) = (cos x)' = -sin x.

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**.

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

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