Question 192947
*[Tex \LARGE f(x)=\left\{\begin{array}{ll} 3x+5 & \ x < -2 \\ x+2 & \ x \ge -2 \end{array}\right.]


Remember, a piecewise function is simply a function composed of different functions. These functions are defined on "pieces" of the domain. So in the case of the given function 


*[Tex \LARGE f(x)=\left\{\begin{array}{ll} 3x+5 & \ x < -2 \\ x+2 & \ x \ge -2 \end{array}\right.]


the two equations {{{y=3x+5}}} and {{{y=x+2}}} are defined for {{{x<-2}}} and {{{x>=-2}}} respectively. So what this means is that you graph the equation {{{y=3x+5}}} ONLY if {{{x<-2}}}. Otherwise, you graph {{{y=x+2}}} (since it is defined for {{{x>=-2}}})



So the graph will look like this:

 
{{{ graph( 500, 500, -10, 10, -10, 10, (3x+5)(sqrt(-x-2)/sqrt(-x-2)),(x+2)(sqrt(x+2.2)/sqrt(x+2.2))) }}}


Graph of the given piecewise function where the red line is {{{y=3x+5}}} (only drawn if {{{x<-2}}}) the green line is {{{y=x+2}}} (only drawn if {{{x>=-2}}})


Note: there is an open circle at (-2,-1) and a closed circle at (-2,0).



So because {{{-2>=-2}}}, this means that we'll use {{{y=x+2}}} to find {{{f(-2)}}}


Because the function {{{f(x)=x+2}}} is defined for values {{{x>=-2}}}, this means that {{{f(-2)=-2+2=0}}}



{{{f(x)=x+2}}} Start with the second expression of the piecewise function



{{{f(-2)=-2+2}}} Plug in {{{x=-2}}}



{{{f(-2)=0}}} Add



So {{{f(-2)=0}}}