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My book says:

If P(x) and Q(x) are polynomials, then the inequalities regarding $\frac{P(x)}{Q(x)}$ can be solved by ALGORITHM:

$(1)$Plot the critical points on number-line. Note $n$ critical points will divide the number line in $n+1$ regions.

$(2)$ In the right most region, the expression will be positive and in other regions it will be alternatively negative and positive. So, mark positive sign int the right most region and than mark alternatively negative and positive signs in the remaining regions.

I have seen some examples following and contradicting the algorithm. some of the examples are:

$(1)$Example which follows the algorithm: $$f(x)=\frac{4x^2+1}{x}>0$$ In the example I get critical points $-1/2$ and $1/2$. Number line so our inequality is positive on $(-\infty,\frac{1}{2})\cup(\frac{1}{2},\infty)$

$(2)$ Example which contradicts :$$f(x)=(\frac{x}{2+x})^2\frac{1}{1+x}>0$$ here I get critical points $-1$,$0$. But the Algorithm doesn't give me correct result.Numberline The actual solution is:$(-1,0)\cup(0,\infty)$ here we neglected $zero$ than ploted the regions, but WHY?

$(3)$In another case, $zero$ is a solution and we take into account $zero$ when plotting regions.$$f(x)=x^4-2x^2>0$$ solution is: $(-1,0)\cup(1,\infty)$

Someone Please Explain. the algorithm and where am i wrong in understanding my book.

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    sorry I couldn't upload picture for third equation as i didn't have that much reputations to do that. – 2017-02-05
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    In your first example, $f(x)$ is negative in $(-\infty,1/2)$. The critical points of $Q(x)$ can also change the sign of $f(x)$. What happens at roots of multiplicity greater than $1$? For example, $f(x)=x^2$? – 2017-02-05
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    so what does that algorithm in my book means? did i understood it wrong? – 2017-02-05
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    The algorithm you have written is nonsense without further assumptions on $P(X)$ and $Q(X)$. I am unsure what you even mean by "inequalities can be solved"... – 2017-02-05
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    I can't post the picture of the page of book until i get 10 reputations. – 2017-02-05
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    If you mean "finding what regions the rational function $f(x)$ is positive", there is an algorithm you can use but it involves slightly more than the above. – 2017-02-05
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    yes yes...i mean that only – 2017-02-05
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    OK, I will write the general method as an answer. Also in $(2)$ above we exclude $0$ since the inequality is strict. $f(0)=0$ and so we exclude it. – 2017-02-05
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    ok fine, but why would you exclude zero in $(2)$, it isn't a solution? – 2017-02-05
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    ...is $f(0)=0>0$? – 2017-02-05
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    oh, okay.....fine then. – 2017-02-05

1 Answers 1

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Here is a method to determine which regions a rational function $f(x)=\frac{P(x)}{Q(x)}$ is strictly positive.

  1. First you find the critical points $P(x)$ and $Q(x)$ and corresponding multiplicities.
  2. Break the number line up by the critical points you found in $1$.
  3. Look at the leading coefficients of $P(x)$ and $Q(x)$. If they share the same sign then $f(x)$ is positive in the right most region, otherwise is is negative.
  4. Working from right to left, once you go over a critical point, if the multiplicity is odd, then the sign of $f(x)$ changes. If it is even, then the sign of $f(x)$ remains unchanged. If it is a critical point of $Q(x)$ and even we also include the critical point in the region.

Note, if $P(x)$ and $Q(x)$ share a critical point, do as above but look at the difference in multiplicities to determine if odd or even.

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    could you provide any link for detailed algorithm with examples? – 2017-02-05