



 
Hi Grayson, I want to show you a way to approach this problem and others like it. I am going to consider the function \[g(x) = x^4 + 2x^3  3x^2\] and ask for what intervals of the real line is $g(x)$ positive and for what intervals is it negative? As a first step I want to find all real numbers $x$ for which $g(x) = 0.$ To do this I ned to factor $g(x).$ \[g(x) = x^4 + 2x^3  3x^2 = x^2\left(x^2 + 2x  3\right) = x^2(x1)(x+3).\] Hence $g(x) = 0$ at $x = 3, 0$ and $1.$ These three points subdivide the real line into four intervals,
Next select a point in one of these intervals, I am going to select 1 which is in the interval from 3 to 0. $g(1) = 4$ hence $gx)$ is negative at $x = 1.$ Can there be an $x$ in the interval from 3 to 0? The fact that $g(x)$ is continuous means that the answer is no. This is a consequence of the Intermediate Value Theorem. If $g(a)$ is positive and $g(b)$ is negative then for some $x$ between $a$ and $b, g(x) = 0.$ but $g(x)$ is only 0 at $x = 3,0$ and $1.$ Hence $g(x)$ is negative on the interval from 3 to 0. Now try a different point in a different interval, say $x = 2$ in the interval from 1 to plus infinity. Is $g(2)$ positive or negative. Whatever sign it has it retains for the entire interval. Finally look at the remaining two intervals. To verify your answer here is a graph of $y = g(x).$ I hope this helps,  


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