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:
its c
Step-by-step explanation:
Answer:
(f o g)(4)=27
Step-by-step explanation:
(f o g)(4)=2x+5
f(g(4))=2x+5
f(4²-2(4)+3)=2x+5
f(16-8+3)=2x+5
f(8+3)=2x+5
f(11)=2x+5
f(11)=2(11)+5
f(11)=22+5
f(11)=27
Therefore, (f o g)(4)=27
Answer:
8.025
Step-by-step explanation:
If you count the zeros in 1000, it's three. Count three starting from the back of 25. Add 0 in front of 25 to make it a three digit number. You then put the decimal point with the whole number.
