Question 1200105
<pre>
{{{matrix(1,3, "f(x)",""="",

system(

matrix(3,3,sin(3x)/x, if, x<0, a, if, x=0, (5-sqrt(25-x))/(bx), if, x>0)))}}}

The 3 parts of the graphs are 

{{{y=sin(3x)/x}}}, {{{y=a}}}, {{{y=(5-sqrt(25-x))/(bx)}}}

The first function must approach the same value as the second function as x approaches 0 from the left. So

{{{lim["x->0-"]}}}{{{(sin(3x)/x)}}}{{{""=""}}}{{{lim["x->0"]}}}{{{3*(sin(3x)/(3x))}}}{{{""=""}}}{{{3*lim["x->0"]}}}{{{(sin(3x)/(3x))}}}{{{""=""}}}{{{3*(1)}}}{{{""=""}}}{{{3}}}

Therefore a = 3, so the second function is y=3

Also the third function must approach the same value as the second function as
x approaches 0 from the right.

{{{lim["x->0+"]}}}{{{((5-sqrt(25-x))/(bx))}}}{{{""=""}}}{{{lim["x->0+"]}}}{{{(((5-sqrt(25-x))/(bx))((5+sqrt(25-x))/(5+sqrt(25-x)))^"")}}}{{{""=""}}}
{{{lim["x->0+"]}}}{{{((25-(25-x))/(bx(5+sqrt(25-x)))))}}}{{{""=""}}}{{{lim["x->0+"]}}}{{{(25-25+x)/(bx(5+sqrt(25-x)))))}}}{{{""=""}}}{{{lim["x->0+"]}}}{{{(x)/(bx(5+sqrt(25-x)))))}}}{{{""=""}}}
{{{lim["x->0+"]}}}{{{cross(x)/(b*cross(x)(5+sqrt(25-x)))))}}}{{{""=""}}}{{{lim["x->0+"]}}}{{{1/(b*(5+sqrt(25-x)))))}}}{{{""=""}}}{{{1/(b*(5+sqrt(25-0)))))}}}{{{""=""}}}{{{1/(b*(5+sqrt(25)))))}}}{{{""=""}}}{{{1/(b*(5+5))))}}}{{{""=""}}}{{{1/(10b)}}}

So {{{1/(10b)}}}{{{""=""}}}{{{3}}}

{{{1}}}{{{""=""}}}{{{30b}}}

{{{1/30}}}{{{""=""}}}{{{b}}}

Here's the graph of the continuous function:

{{{drawing(400,250,-8,8,-5,5,

graph(400,250,-8,8,-5,5, (5-sqrt(25-x))/((1/30)x)*(sqrt(x)/sqrt(x))),

graph(400,250,-8,8,-5,5, (sin(3x)/x)(sqrt(-x)/sqrt(-x)))





    )}}}

Edwin</pre>