Question

Can anyone help me with this ASAP?

Answer 1

qwertyuiop qwertyuiop

Explanation:

Answer 2

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

`(x-3)/(x-3)=1`

giving

`(x(x+2)^2(x-\pi)^3(x-3))/(x^2(x-3)^2(x+\pi)^4) \ge 0`

Everything in the denominator is positive now, so we can clear it, multiplying it on both sides. It vanishes on both sides.

`x(x+2)^2(x-\pi)^3(x-3) \ge 0`

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.

`x(x-\pi)(x-3) \ge 0`

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.]