Question 176583
I'm assuming the brackets signify absolute value??
Break the problem down into 3 distinct regions based on the arguments of the absolute value functions.
{{{3x+7<0}}}
{{{3x<-7}}}
{{{x<-7/3}}}
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{{{1-x<0}}}
{{{-x<-1}}}
{{{x>1}}}
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and in between these two regions.
{{{-7/3<x<1}}}
In the region where x<-7/3
{{{abs(1-x)=(1-x)}}}
{{{abs(3x+7)=-(3x+7)}}}
The function then becomes
{{{G(x)=(1-x)-(-(3x+7))}}}
{{{G(x)=(1-x)+(3x+7))}}}
{{{G(x)=2x+8}}}
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In the region where x>1, then
{{{abs(1-x)=-(1-x)}}}
{{{abs(3x+7)=(3x+7)}}}
The function then becomes
{{{G(x)=-(1-x)-(3x+7)}}}
{{{G(x)=-1+x-3x-7))}}}
{{{G(x)= -2x-8}}}
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Finally in between these two regions.
{{{-7/3<x<1}}}
{{{abs(1-x)=(1-x)}}}
{{{abs(3x+7)=(3x+7)}}}
The function then becomes
{{{G(x)=(1-x)-(3x+7)}}}
{{{G(x)= -4x-6))}}}
{{{drawing( 300, 300,  -5, 4, -15, 10,grid( 1 ),
circle( -7/3, 10/3, .2 ), 
circle( 1, -10, .2 ), 
graph( 300, 300, -5, 4, -15, 10, 2x+8,-2x-8,-4x-6)) }}}
Red : G(x)=2x+8
Blue :G(x)= -4x-6
Green :G(x)=-2x-8
The circles show the break points of x=-7/3 and x=1.
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Here is the graph trimmed of the extraneous regions.
{{{ graph( 300, 300, -5, 4, -15, 10, abs(1-x)-abs(3x+7)) }}}