



 
Rocco, I'm going o look at a somewhat similar but simpler problem.
I see this in a graphical way. I imagine that I am walking along the curve f(x) = x^{3}  8x^{2} + 16x and ask when I am above the line y = 0. The key is to ask when might I cross from one side of y = 0 to the other. If I cross from one side of y = 0 to the other I must go through a point where y = 0, that is a point where
To find such points I factor x^{3}  8x^{2} + 16x and obtain
This has solutions x = 0 and x = 4 and these are the only possible xvalues where I might cross x = 0. Now suppose I am at a point on the graph with xcoordinate less than 0, say for example x = 10. Then f(10) = 10(10  4)^{2} = 1960 so I am below y = 0. The first place I can cross y = 0 at x = 0 so where am I when x = 1? f(1) = 1(14)^{2} = 9 so I am above y = 0. The next place I can cross y = 0 is at x = 4 so where am I if x > 4, say x = 5? f(5) = 5(54)^{2} = 5 so I am still above y = 0 and I can't cross y = 0 again. Thus the answer to
is for 0 < x < 4 and for x > 4. Here is the graph . For your problem
first solve (x  2)/(x + 4) = 7 to see where the graph of f(x) = (x  2)/(x + 4) crosses the line y = 7. You should find one xvalue. The complication in your problem is what happens at x = 4. At this point the denominator of f(x) = (x  2)/(x + 4) is zero and there is no point on the curve. The graph near x = 4 might look something like and the curve might then move from one side of y = 7 to the other without actually crossing y = 7. Thus, as in my problem you have two xvalues where the graph might move from one side of y = 7 to the other, the xvalue you found by solving (x  2)/(x + 4) = 7 and x = 4. Continue as I did for the simpler problem. Write back if you need further assistance. Harley  


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