Trig Help - Inequalities, critical numbers

[dragonflame327]

I took the ASVAB today, so I missed class. I talked to my teacher and he said the homework was just what we've been doing, which is solving inequalities using critical numbers. On the homework though, quartic equations are showing up, which he must've covered in class, because I keep getting them wrong. I checked the answer in the back of the book and it's not matching up with what I get.

The first problem that I did:

x^3 - 3x^2 - x +3 > 0 I'd seperate these and treat them like seperate equations, right? So...

(x^3 -3x^2)( -x +3) > 0
x^2(x - 3)(-x + 3)
x = 0, and 3

The book says the solution (I'm skipping the step where we use a number line to aid us.), is (-1, 1)U( 3, infinity)

What am I doing wrong?

Also, I'm looking ahead at other problems I have to do tonight...

4x^3 - 6x^2 < 0 Woul I do it like this?
2x^2(2x-3)
x= sqrt(2), -3/2

And then I'd just treat it like I would the critical numbers of a quadratic? Or am I doing it wrong?

Any help would be much apprieciated. Thank you.^^

 

[stapel]

Note: A "quartic" is a fourth-degree polynomial; a third-degree polynomial is a "cubic".

x^3 - 3x^2 - x +3 > 0
I'd seperate these and treat them like seperate equations, right?

If you mean that you would factor the polynomial, then, yes, this would be the first step. But you don't "treat them like separate equations". Instead, you would find the zeroes (you would treat the factors separately for this), use these zeroes to divide the number line into intervals, and then test the sign of the polynomial on each interval.

(x^3 -3x^2)( -x +3) > 0

If you multiply this out, you would get a quartic, not the original cubic.

It would appear that you have become accustomed to being overly casual in your notation and have reached the point where imprecision is now giving you the wrong answers. The cubic factors as:

. . .x³ - 3x² - x + 3

. . . . .= x²(x - 3) - 1(x - 3)

. . . . .= (x - 3)(x² - 1)

Find the zeroes. Then test the original polynomial between the zeroes. The intervals on which the polynomial returns a positive value constitute the solution.

4x^3 - 6x^2 < 0 Woul I do it like this?
2x^2(2x-3)

What happened to the inequality sign?

x= sqrt(2), -3/2

Assuming you have set the factors equal to zero to find the interval endpoints, how are you getting that "x² = 0" means x = sqrt(2)?

Eliz.