Function Domain Help: Solving x^2-x-6>=0

[lukatwo]

Homework Statement

Hello! Could someone explain some things presented in this example: http://i.imgur.com/wVOxf.png
My friend, and I haven't been able to crack it. I get all the x-conditions right, but can't put the whole thing together into a domain. Here it's presented that xe[-2,3] due to the positive values being between them. That confuses me since i thought that when the quadratic equation has a positive index the graph of the parabola should be facing up with the opening.
Sorry for my bad math jargon(ESL)

Homework Equations

x^2-x-6>=0

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

Hello! Could someone explain some things presented in this example: http://i.imgur.com/wVOxf.png
My friend, and I haven't been able to crack it. I get all the x-conditions right, but can't put the whole thing together into a domain. Here it's presented that xe[-2,3] due to the positive values being between them. That confuses me since i thought that when the quadratic equation has a positive index the graph of the parabola should be facing up with the opening.
Sorry for my bad math jargon(ESL)

Homework Equations

x^2-x-6>=0

The Attempt at a Solution

If you are trying to figure out where$$
\frac {x^2-x-6}{x-7}\ge 0$$
your answers of $[-2,3]$ and $(7,\infty)$ are correct.

 

[lukatwo]

Could you explain why? I understand the lower part, but the quadratic inequation bothers me. The first index is a>0 => parabola with a minimum vertex, meaning it's values are >=0 from <-inf,-2]U[3,+inf>. That is the reason i made this thread.

 

[Villyer]

Could you explain why? I understand the lower part, but the quadratic inequation bothers me. The first index is a>0 => parabola with a minimum vertex, meaning it's values are >=0 from <-inf,-2]U[3,+inf>. That is the reason i made this thread.

That is true if it is just the quadratic equation, but if you are dividing two numbers what happens if they are each positive? Each negative? One positive and one negative?
You need to use the answers to those questions to determine the sign of the division of two functions.

 

[LCKurtz]

Could you explain why? I understand the lower part, but the quadratic inequation bothers me. The first index is a>0 => parabola with a minimum vertex, meaning it's values are >=0 from <-inf,-2]U[3,+inf>. That is the reason i made this thread.

Look at the signs of the factors in$$
\frac{(x-3)(x+2)}{x-7}$$on those intervals. They determine the sign of the function.

 

[lukatwo]

That is true if it is just the quadratic equation, but if you are dividing two numbers what happens if they are each positive? Each negative? One positive and one negative?
You need to use the answers to those questions to determine the sign of the division of two functions.

I did take that in account, and made 2 scenarios (unlike the example):
1.positive or zero/positive
2.negative or zero/negative

In the first case i get <-inf,-2]U[3,+inf>
and <7,+inf> !

In the second case i get [-2,3] !
and <-inf,9>
So how do i get half the right answer in 2 different scenarios?

 

[eumyang]

For questions like these I like to create a sign chart, which is essentially a number line. First you need to find values of x that make the rational expression 0 and undefined. These form your critical values that you can label on a number line like this:

         0        0       und
<--------+--------+--------+-------->
        -2        3        7

You have 4 intervals to test: (-∞, -2), (-2, 3), (3, 7), and (7, ∞). Test values of x within each interval to determine the sign of the rational expression. For x in (-∞, -2), all binomial factors are negative, so the entire expression is negative, so I label the sign chart as thus:

  (-)(-)
  ------
   (-)   0        0       und
<--------+--------+--------+-------->
   neg. -2        3        7

The "(-)"s indicate negative binomials within the rational expression.
Repeat the process with the other 3 intervals. You solution set will be the values that make the entire expression positive or 0.

 

[LCKurtz]

Look at $$
\frac{(x-3)(x+2)}{x-7}$$
There are 3 factors, and each factor changes sign at its root. For example, for $x>7$ all three factors are positive, so the signs are $\frac {++}{+}$. Think of $x$ slowly decreasing along the number line. When $x$ becomes less than 7 the $x-7$ factor changes sign so the signs are $\frac {++}{-}$. It stays that way until $x$ crosses 3 at which point the factor $x-3$ changes sign so you have $\frac {-+}{-}$. Finally when $x$ passes -2 the factor $x+2$ becomes negative and the signs are $\frac {--}{-}$. And, of course, the fraction is 0 when $x=3$ or $x=-2$.

 

[lukatwo]

Thank you very much eumyang and LCKurtz! This really helps a lot. Now i realize that i have to look at it as a whole, and not two separate inequations. I don't know if there's some repping system on this forum but you definitively get my thanks/+1