Step-by-step explanation:
please the information needed is up the paper. you took a picture of no. 2 instead of no. 1

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

Hayden had the membership for one month and took 6 classes.
3a + 6b = 45
2a - 2b = -12....multiply by 3
----------------
3a + 6b = 45
6a - 6b = - 36 ...(result of multiplying by 3)
---------------add
9a = 9
a = 1
3a + 6b = 45
3(1) + 6b = 45
3 + 6b = 45
6b = 45 - 3
6b = 42
b = 7
solution is : (1,7)