Posted by justme on Saturday, September 24, 2011 at 10:52am.
if we rename the two functions f and g, then g(x) = f(x+3) - 1
So, the graphs are identical in shape, but g is shifted two units to the left, and up 1.
So, since f(x) = x^3 (x+1) the zeroes are at 0,-1
Now, if we draw two new axes, and call them v and v, then the zeroes of g are the solutions to u^4 + u^3 - 1. That's a little tougher to solve.
A graphing tool will show the approximate ranges. Domain is, of course, all real numbers.
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