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# For the function g(x)= x-2/x+4 , solve g(x) < 0. please show the work.

For the function g(x)= x-2 / x+4 , solve g(x) < 0 . Please make sure you show the work

QUESTION: g(x)=(x-2)/(x+4) ; solve g(x)<0

SOLUTION:
The numerator of this ratio is (x-2); the denumerator is (x+4).

Let start solving separately the two simple inequalities (x-2)>0 and (x+4)>0.

Numerator : (x-2)>0 means x>2;

Denominator: (x+4)>0 means x>-4.

So, in this way we know that the numerator is positive when x is greater than 2 while the denominator is positive when x is greater than -4.

Asking to solve g(x) <0 means that the quotient has to be negative. That could be only if "the numerator is positive and the denominator is negative" or viceversa "the numerator is negative and the denominator positive".

Let me use a graph:
-4                    2
------|----------------|-------------- real line
Num.       -             -                 +

Den.        -            +                 +

Is it clear now that the solution is the set of number between -4and 2, that is the open interval (-4;2).

g(x)<0 in this case is:

(x-2)/(x+4)<0;

The ratio of two numbers is less than zero when one is positive and another one is negative.

So we have two cases:

x-2<0 and
x+4>0

OR

x-2>0 and
x+4<0

In the first case,
x<2 and
x>-4,
which is equivalent to -4<x<2 or x∈(-4;2)

In the second case,
x>2 and
x<-4
There is no way that x can be simultaneously greater than two and less than -4, so there is no solution for the second system of equations. Or it is said that the solution is the empty set of points. Anyway, the solution to the inequality is the union of two sets: the solution to the first case and the solution to the second case. Since the second case has no solution, the union is just the set of x for the first case.