We need to solve the following inequality:
[tex]\frac{9-x}{x+11}\ge0[/tex]Then we have that for the inequality would be complied we have two implicit conditions:
[tex]9-x\text{ }\ge\text{ 0}[/tex]And at the same time:
[tex]x+11>0[/tex]You have to be careful because we already know that the denominator of a fraction can not be zero, it's, for this reason, the second inequality.
But, in a second case, we can also have both numerator and denominator as negative numbers, it also gives us a number bigger or equal to zero.
So we have the inequalities:
[tex]9-x\leq0[/tex]And:
[tex]x+11<0[/tex]Firstly we can focus on the first case if we solve for x:
[tex]9-x\ge0[/tex][tex]x\leq9[/tex]And the denominator inequality of this case:
[tex]x+11>0[/tex][tex]x>-11[/tex]And how we must have the agreed interval between the conditions, we have that the first result for this case is the interval:
(-11,9], or in a equivalent form: -11
From the second case, when both numerator and denominator we have:
[tex]9-x\leq0[/tex][tex]x\ge9[/tex]And from the denominator inequality:
[tex]x+11<0[/tex][tex]x<-11[/tex]So a second result is an interval that doesn't exist because a number biggest of 9 and smallest than -11 doesn't exist in the real number.
Then we obtain the final result, and the correct answer is:
(-11,9], or in a equivalent form: -11
D.
