What's with the gratuitous pi s?
The rule is to preserve inequalities we can only multiply both sides by positive numbers, or non-negative numbers (zero ok) for ≤ and ≥ (but not < and >).
To clear the denominator, first we need all positive factors down there so we multiply by
[tex]\dfrac{x-3}{x-3}=1[/tex]
giving
[tex]\dfrac{x(x+2)^2(x-\pi)^3(x-3)}{x^2(x-3)^2(x+\pi)^4} \ge 0[/tex]
Everything in the denominator is positive now, so we can clear it, multiplying it on both sides. It vanishes on both sides.
[tex]x(x+2)^2(x-\pi)^3(x-3) \ge 0[/tex]
Again we can divide both sides by the positive factors, which are the squares, including (x-π)². We'll get zero when those factors are zero (and no zero denominator factor), so we have to remember to include x=0, x=-2, x=3 and x=π in the answer, but we eliminate 0 and 3 from this list because they give zero denominators.
[tex]x(x-\pi)(x-3) \ge 0[/tex]
That's a lot easier to think about. When x is really negative we're multiplying three negative numbers, so a negative result.
When x is really big we're multiplying three positive numbers, again a positive result.
Between 0 and pi it's going to alternate.
<0 negative
0-3 positive
3-π negative
>π positive
Since zero is ok we can use ≤ or ≥ as appropriate, but we should also rule out anything that gives a zero denominator, as the truth value of 'undefined ≥ 0' is not really determined.
Answer: 0 < x < 3 or x ≥ π or x=-2
[I missed the x=-2 the first time around; sorry about that.]
