Answer:
6.7 I think
Step-by-step explanation:
hope this helps
Answer:
ummm I think this is obvious
Step-by-step explanation:
2 circles in each group, shade in one group of 2.
To check for continuity at the edges of each piece, you need to consider the limit as
approaches the edges. For example,
![g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}](https://tex.z-dn.net/?f=g%28x%29%3D%5Cbegin%7Bcases%7D2x%2B5%26%5Ctext%7Bfor%20%7Dx%5Cle-3%5C%5Cx%5E2-10%26%5Ctext%7Bfor%20%7Dx%3E-3%5Cend%7Bcases%7D)
has two pieces,
and
, both of which are continuous by themselves on the provided intervals. In order for
to be continuous everywhere, we need to have
![\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto-3%5E-%7Dg%28x%29%3D%5Clim_%7Bx%5Cto-3%5E%2B%7Dg%28x%29%3Dg%28-3%29)
By definition of
, we have
, and the limits are
![\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto-3%5E-%7Dg%28x%29%3D%5Clim_%7Bx%5Cto-3%7D%282x%2B5%29%3D-1)
![\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto-3%5E%2B%7Dg%28x%29%3D%5Clim_%7Bx%5Cto-3%7D%28x%5E2-10%29%3D-1)
The limits match, so
is continuous.
For the others: Each of the individual pieces of
are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.
USE SOCRACTIC IT WOULD REALLY HELP WITH THIS QUESTION