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 .
The Solution:
We are given the decimal below:
We are asked to express the given decimal as a fraction in its lowest term.
Therefore,
Answer:
Im not sure
Step-by-step explanation:
GOOD LOOK ON YOUR TEST HE DISCONN-ECTED
Answer:
vertical asymptote X=-1
horizontal asymptote Y=2
Step-by-step explanation: