Continuity and Finding Values Find value of a,b,c such that F is continuous on the real number system:
$$
  f(x) = \left\{
    \begin{array}{lr}
      -1 & : x\le-1\\
      ax^2+bx+c & : |x|<1,x\ not\ equal\ to\ 0\\
      0 &:x=0\\
      1 &:x\ge1
    \end{array}
  \right.
$$
We went over the following solution in class:
Show $f(x)$ is continuous at -1:
For all $\epsilon > 0, \exists\ \delta$ such that for all $x, |x+1|$ < $\delta$ and $|f(x)-f(-1)|=$
if $x\le-1: f(x)=f(-1)=-1$ and $|f(x)-f(-1)|=0<\epsilon$
if $x>-1: |f(x)-f(-1)|=|ax^2+bx+c-(-1)|=|a[x^2-(-1)^2]+b[x-(-1)]+c-(-1)+a(-1)^2+b(-1)|<\epsilon.$
so $a-b+c+1=0$
Show $f(x)$ is continuous at 0:
For all $\epsilon > 0, \exists\ \delta\ $ such that for all $x, |x-0|<\delta\ $ and $|f(x)-f(0)|=|f(x)|$=$|ax^2+bx+c| = x$
$c=0$$a+b+c-1=0$
$a-b=-1,a+b=1$, thus $a = 0$ and $b = 1$I am having difficulty processing through this solution. I am first confused with why for showing the continuity at -1, it was broken down into two cases ($x\le-1$ and $x>-1$).
Then, how did we simplify into $a-b+c+1=0$ in that second case?Finally, in showing continuity at $0$, how did we know that $c = 0$?
 A: 
I am first confused with why for showing the continuity at -1, it was broken down into two cases ($x\le−1$ and $x>−1$).

Because the function is defined differently for $x\le-1$ and $(|x|<1, x\neq0)$ so in order to think about its continuity it's useful to see how it behaves either side of -1.

Then, how did we simplify into $a−b+c+1=0$ in that second case?

Well, the $x\le -1$ bit was clearly always continuous, so forget about that.
For the other bit, note that rewriting the difference as $$|a[x^2−(−1)^2]+b[x−(−1)]+c−(−1)+a(−1)^2+b(−1)|$$ makes sure that the brackets involving $x$ both go to $0$ as $x\to-1$. If we want to be able to choose a $\delta$ to make this whole thing less than $\epsilon$, then the first two brackets will always go to zero, leaving us with $|[\text{small}]+c+1+a-b|$. Clearly if this is going to go to be less than all small $\epsilon$ then we must have $\boxed{c+1+a-b=0}$ (because it doesn't depend on $x$).
It's a lot easier to see this a difference way though. Consider the limit of $f(x)$ as $x\to-1$ from above:
$$\lim_{x\to -1^+}f(x) = \lim_{x\to -1^+}(ax^2+bx+c) = a-b+c$$
This must be equal to $f(-1)=-1$, so $\boxed{a-b+c=-1}$.

Finally, in showing continuity at 0, how did we know that $c=0$?

(You have an odd looking $=x$ at the end of the line in the workings for this bit.)
Near $x=0$, we know that $f(x)=ax^2+bx+c$, whilst $f(0)=0$. Again, for small $x$ we have $$|f(x)-f(0)|=|ax^2+bx+c|\to |c|$$
Again, this must be zero if we have continuity.

In general, it's easiest to think about continuity at $y$ in terms of $\lim_{x\to y} f(x) = f(y)$.
A: The reason they break it down into two cases at $x=-1$ is that the function is fundamentally different on $\{x<-1\}$ than it is on $\{x>-1\}$.
The way they simplify into $a-b+c+1=0$ is by simply plugging $x=-1$ into $ax^2 + bx + c - (-1)$.
The way we knew that $c=0$ is that as $x$ gets small, $ax^2 + bx + c$ will get arbitrarily close to $c$. If $f$ is continuous, it also must get arbitrarily close to $f(0) = 0$. Therefore, $c=0$.
