Is this function continuous everywhere? is $f(x)= |3-x|+|x+2|$
I graphed it using Desmos and Wolfram and its looks like it's continuous everywhere. I just don't know how to prove this without using a graphic program.
Does it have to do with the absolute values?
 A: on the real line $|3-x|= \max(3-x,x-3)$
the max of two continuous functions is continuous.
similarly for $|x+2| = \max(x+2, -x-2)$
now you can use the fact that the sum of two continuous functions is continuous.
A: So basically, you probably already know that $f(x) = 3 - x$, which is a line, is continuous everywhere.
Similarly, $g(x) = x+ 2$, which is also a line, is continuous everywhere.
Now you need to know that if you have a function continuous everywhere, the absolute value of that function is continuous everywhere.  So $h(x) = |3 - x|$ is continuous everywhere since it is the absolute value of a function that is continuous everywhere.
Similarly, $k(x) = |x + 2|$ is continuous everywhere since it is the absolute value of a function that is continuous everywhere.
Finally, you need to know that if you have two functions that are continuous everywhere, then when you add them, that new function is continuous everywhere.  So since $|3 - x|$ and $|x + 2|$ are continuous everywhere, so is $|3 - x| + |x + 2|$.
Because of everything above, we can say $t(x) = |3 - x| + |x + 2|$ is continuous everywhere.
A: It is continuous.
The sum of two continuous function is still continuous and as we know $f(x)=|g(x)|$ is continuous if $g(x)$ is continuous.
