For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that 
is such that
Obviously, both limits are 0, so 
is indeed continuous at 
.
Now, for to be differentiable everywhere, its derivative 
must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:
and at this time, we don't know what's going on at , so we omit that case. We want to be continuous, so we require that
from which it follows that 
.
Answer:
Your answer would be 70.434! Sorry if its wrong, but im sure its right!
I'd wait for somebody else to answer just to confirm!
That would be C....(1 x 10^-1) + (6 x 10^-2) + (9 x 10^-3)
Answer:
18÷2 × 3 / 5-2
9 ×3 / 3
27/3
9
tyvm this was my 100th answer :)
