Question 1020442
When {{{x>=1/3}}},
{{{abs(3x-1)=3x-1}}}
So then,
{{{f(x)=2x+3x-1=5x-1}}}
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When {{{x<1/3}}},
{{{abs(3x-1)=-(3x-1)=1-3x}}}
So then,
{{{f(x)=2x+1-3x=1-x}}}
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So the critical point is {{{x=1/3}}}, where the slope of the line goes from negative to positive. 
So the function must be a minimum at that point.
{{{f(1/3)=5(1/3)-1=5/3-1=2/3}}}