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If f(x)=x(x+3)(x-1), use interval notation to give all values of x where f(x)>0.
\n" ); document.write( "Find the places that the function = 0, this will tell us the intervals to test.
\n" ); document.write( "0=x(x+3)(x-1)
\n" ); document.write( "x=0 and x+3=0 and x-1=0
\n" ); document.write( "x=0 and x=-3 and x=1
\n" ); document.write( "The intervals are (-infinity,-3),(-3,0),(0,1), and (1,infinity)
\n" ); document.write( "For (-infinity,-3) test x=-4
\n" ); document.write( "-4(-4+3)(-4-1)>0 ?
\n" ); document.write( "-4(-1)(-5)>0 an odd number of negatives gives us a negative
\n" ); document.write( "-20>0 is false, so this interval is not part of the solution set.
\n" ); document.write( ":
\n" ); document.write( "For (-3,0), test x=-1
\n" ); document.write( "-1(-1+3)(-1-1)>0 ?
\n" ); document.write( "-1(2)(-2)>0 an even number o0f negatives is positve
\n" ); document.write( "4>0 is true, so (-3,0) is part of the solution set.
\n" ); document.write( ":
\n" ); document.write( "For (0,1) test x=1/2
\n" ); document.write( "1/2(1/2+3)(1/2-1)>0?
\n" ); document.write( "1/2(7/2)(-1/2)>0 an odd number of negatives gives a negative.
\n" ); document.write( "-7/8>0 is false, so this interval is not part of the solution.
\n" ); document.write( ":
\n" ); document.write( "For (1,infinity) test x=2
\n" ); document.write( "2(2+3)(2-1)>0?
\n" ); document.write( "2(5)(1)>0 no negatives make a postive
\n" ); document.write( "10>0 is true, so (1,infinity) is part of the solution.
\n" ); document.write( ":
\n" ); document.write( "Therefore the solution is: (-3,0)U(1,infinity)
\n" ); document.write( "Happy Calculating!!! \n" ); document.write( "