Solve the following inequality.

x^2(6+x)(x-6)/(x+2)(x-2) ≥0

Question

Solve the following inequality. 

x^2(6+x)(x-6)/(x+2)(x-2)  ≥0

anonymous · Answer

@whpalmer4

whpalmer4 · Answer

is that $$\frac{x^2(6+x)(x-6)}{(x+2)(x-2)} \ge 0$$

anonymous · Answer

yes

whpalmer4 · Answer

Hmm...let me think about how to do this correctly...

anonymous · Answer

ok =)

whpalmer4 · Answer

Well, first thing I'll observe is that the function is undefined at $x = \pm 2$ because the denominator becomes 0 at those two values.

whpalmer4 · Answer

Similarly, it's going to be 0 at $x =0,\,x=\pm 6$ because the numerator is 0 at those values of $x$.

Agreed?

whpalmer4 · Answer

From x = 2 and greater, the denominator is positive.

anonymous · Answer

ok

whpalmer4 · Answer

Now, we need to figure out what range for x=2 and greater will also have a positive numerator, since to get the result >= 0, we want the numerator to be positive when the denominator is positive.

If we just look at the numerator:$$x^2(x+6)(x-6)$$That's positive if $x\ge 6$
At $x=0$, the numerator is 0, so that's also in our solution
From $0 < x < 6$ the numerator is negative, because of the $x-6$ product term

Agreed with all of that?

anonymous · Answer

yes =)

whpalmer4 · Answer

Ah, but I was wrong, and you shouldn't have agreed so quickly :-)

From $ 0 \le x \le 2$ the numerator and the denominator are both negative, so that gives us a positive quotient!

for example, at x = 1
$$\frac{(1)^2(1+6)(1-6)}{(1+2)(1-2)} = \frac{1*7*-5 } {3*-1 } = \frac{35}{3}$$

So I believe our answer is (for the right side only) $$0 \le x \lt 2 \,|| \,x \ge 6$$Where $||$ means or

whpalmer4 · Answer

And I think this is symmetrical about the y-axis, so really our answer is
$$x \le -6 \,||\,-2< x < 2\,||\,x \ge 6$$

whpalmer4 · Answer

Now for the moment of truth: I plot the function and see what it looks like :-)

Hey, we were right!

anonymous · Answer

it says to type the answer in interval notation?

whpalmer4 · Answer

Okay, in interval notation I think that would be $$[-\infty,-6],(-2,2),[6,\infty]   $$but I don't spend any time using interval notation, so I make no promises...

anonymous · Answer

looks about right lol

whpalmer4 · Answer

I'm more concerned with finding the right answer, and less with jumping through what I regard as artificial hoops (you solved the problem correctly, but got it wrong because you didn't type it the way I wanted it typed)

anonymous · Answer

ya its dumb.. i agree. can I get your help with just one more, and then i will leave you alone? lol

whpalmer4 · Answer

Unfortunately, I have to go run some errands, but I'll be back later on in the evening, I think.

anonymous · Answer

oh ok =) thank you so much for your help!!

whpalmer4 · Answer

You're welcome!