MHB Finding values for which a function is greater that 0

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To determine where the function g(x) = (x - 1)/(x + 2) is greater than 0, identify the critical points at x = 1 (numerator = 0) and x = -2 (denominator = 0). The function is positive when both the numerator and denominator are either both positive or both negative. This leads to two conditions: x > 1 (both positive) and x < -2 (both negative). The function is undefined at x = -2, negative between -2 and 1, zero at x = 1, and positive for x > 1. Thus, g(x) > 0 for x in the intervals (-∞, -2) and (1, ∞).
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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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