Question 804453
<pre>
{{{"f(x)"}}}{{{""=""}}}{{{system( matrix(3,3,

                                     3x, if, -2<=x<0,
                                   2x^2, if,  0<=x<1,
                                  x^2-3, if,  1<x))}}}

The graph of that function is this:

{{{drawing(200,400,-4,3,-8,6, red(circle(-2,-6,.15),circle(1,2,.15),
circle(-2,-6,.15),circle(-2,-6,.13), circle(-2,-6,.11),circle(-2,-6,.09),
circle(-2,-6,.07),circle(-2,-6,.05),circle(-2,-6,.03),circle(-2,-6,.01),

circle(1,-2,.15),circle(1,-2,.13), circle(1,-2,.11),circle(1,-2,.09),
circle(1,-2,.07),circle(1,-2,.05),circle(1,-2,.03),circle(1,-2,.01)),
graph(200,400,-4,3,-8,6,3x*(sqrt(x+2)/sqrt(x+2))*(sqrt(-x)/sqrt(-x))      ),

 graph(200,400,-4,3,-8,6,2x^2*(sqrt(x)/sqrt(x))*(sqrt(1-x)/sqrt(1-x))),

graph(200,400,-4,3,-8,6,(x^2-3)*( sqrt(x-1)/sqrt(x-1) ) )  )}}}

Solve f(x) = 0
       2x² = 0
        x² = 0
         x = 0


f(0) = 3(0)² = 0

f(x) is continuous at x = 0 because

lim f(x) = lim 3x = 0 
x->0<sup>-</sup>      x->0<sup>-</sup>

lim f(x) = lim 2x² = 0 
x->0<sup>+</sup>      x->1<sup>+</sup>

Therefore

lim f(x) = lim f(x) = lim f(x) = f(0) = 0  
x->1<sup>-</sup>      x->1<sup>+</sup>      x->1

which proves that f(x) is continuous at x=0

However f(x) is not continuous at x=1 because

lim f(x) = lim 2x² = 3 
x->1<sup>-</sup>      x->1<sup>-</sup>

lim f(x) = lim x²-3 = -2 
x->1<sup>+</sup>      x->1<sup>+</sup>

Therefore 

lim f(x) &#8800; lim f(x)  
x->1<sup>-</sup>      x->1<sup>+</sup>

Since f(x) is not continuous at x=1, it is not continuous 
everywhere on its domain.

Edwin</pre>