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$$\frac{x^2(x-1)}{(x+2)(x+3)^3}\le 0$$

I'm not even sure where to start on this inequality. In the numerator, $x=0$ or $x=1$ makes the expression $0$ .

In the denominator, $x$ cannot equal $-2$ and $-3$.

I just don't know where to proceed from here. If anyone could help I would really appreciate it. Thank you

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HINT:

  • $x^2$ and $(x+3)^2$ are always non-negative and we know that $x=0$ is a solution so we need only solve $$\frac{x-1}{(x+2)(x+3)}\le0.$$

  • Consider the cases when $x$ is in each of the intervals $(-\infty,-3)$, $(-3,-2)$, $(-2,1]$.

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  • $\begingroup$ Thanks, just wondering why is 1] inclusive ? $\endgroup$ – J.W. Sep 15 '18 at 14:54
  • $\begingroup$ Because it is defined when $x=1$ but not when $x=-2,-3$ as you have observed. $\endgroup$ – TheSimpliFire Sep 15 '18 at 14:54
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Hint: I would start with $$x^2\geq 0$$ then we have $$\frac{x-1}{(x+2)(x+3)^3}\le 0$$ So we have a fraction of the form $$\frac{a}{bc}\le 0$$

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  • $\begingroup$ What is the significance of that form of fraction? Do I test the intervals between x=1, x=-2, and x=-3? The cube on (x+3) confused me $\endgroup$ – J.W. Sep 15 '18 at 14:59
  • $\begingroup$ Note that $$(x+3)^3=(x+3)^2(x+3)$$ With this fraction you will get every possible case, $$---,++-,-++$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '18 at 15:12
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A slightly shorter way (not to explain, but to use):

Your inequality, on the domain $\mathbf R\smallsetminus\{-2,-3\}$ is indeed equivalent to $$f(x)=\frac{x-1}{(x+2)(x+3)}\le 0 $$ Now, by the rule of signs, this rational function has the sign of the polynomial $$p(x)=(x-1)(x+2)(x+3),$$ which has three simple roots and can change sign only at a root (by the IVT). This means its sign is alternating. Now, considering $\lim_{x\to\infty} p(x)=+\infty$ , you know this sign is + on the rightmost interval. Therefore, we get this table of signs: $$\begin{array}{r}x\\ f(x)\end{array}\;\begin{array}{|*{9}{c}|} \hline -\infty &&-3&&-2&&1&&+\infty\\ \hline &- & \| & + & \| & - & 0 & + \\ \hline \end{array} $$

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