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 $c = 0$?

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, $c=0$.
