Finding values for which a function is greater that 0

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SUMMARY

The function \( g(x) = \frac{x - 1}{x + 2} \) is analyzed to determine the x-values for which \( g(x) > 0 \). The critical points are identified at \( x = 1 \) (where the numerator is zero) and \( x = -2 \) (where the denominator is zero). Testing the intervals \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \) reveals that \( g(x) \) is positive for \( x < -2 \) and \( x > 1 \), while it is negative between \( -2 \) and \( 1 \) and undefined at \( x = -2 \).

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I have a function

$$g(x) = \frac{x - 1}{x + 2}$$.

How would I find all of the x-values for which $g(x) > 0$.

I'm used to doing this for functions that need to be factored but not fractions.
 
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tmt said:
I have a function

$$g(x) = \frac{x - 1}{x + 2}$$.

How would I find all of the x-values for which $g(x) > 0$.

I'm used to doing this for functions that need to be factored but not fractions.
There are other was, but this one is fairly standard for polynomial fractions.

There are two critical points here: when the numerator is 0 and when the denominator is 0. So the critical points are at x = 1 and x = -2. Now you have three intervals to test: [math] ( -\infty , -2 ) [/math], [math]( -2, 1)[/math], and [math](1, \infty)[/math]. See which of these are greater than 0.

-Dan
 
A fraction is positive if and only if the numerator and denominator have the same sign.
$$\frac{x- 1}{x+ 2}> 0$$ if and only if either
1) x-1> 0 and x+ 2> 0. The first inequality gives x> 1 and the second gives x> -2. They will both be true for x> 1.

2) x-1< 0 and x+ 2< 0. The first inequality gives x< 1 and the second x< -2. They will both be true when x< - 2.

The fraction is positive when x< -2, is not defined at x= -2, is negative when -2< x< 2, is 0 when x= 1, and is positive again when x> 1.
 

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