Like what is you trying to say explain

In order to be differentiable everywhere,
must first be continuous everywhere, which means the limits from either side as
must be the same and equal to
. By definition,
, and


so we need to have
.
For
to be differentiable at
, the derivative needs to be continuous at
, i.e.

We then need to have

Then

Answer:
A: 40%
B: 75%
C: 90%
Step-by-step explanation: