91影视

In the following exercises, we have to solve rational inequality and write the solution in interval notation.

$\frac{x+1}{x鈭�3}鈮�0$

Our inequality is in this form.

$\frac{x+1}{x鈭�3}鈮�0$

The rational expression will be zero when the numerator is zero. Since $x+1=0$when $x=-1$, then $-1$is a critical point.

The rational expression will be undefined when the denominator is zero. Since $x-3=0$when $x=3$, then $3$is a critical point.

The critical points are $-1$and $3$.

The number line is divided into three intervals:

$(-鈭�,-1)(-1,3)(3,鈭�)$

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

$\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{鈭�}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$

The number $-2$is in the interval $(-鈭�,-1)$. Test in the $x=-2$expression in the numerator and the denominator.

$\frac{x+1}{x鈭�3}=\frac{-2+1}{-2鈭�3}=\frac{-1}{-5}=\frac{1}{5}$

The quotient positive in the interval $(-鈭�,-1)$.

The number $x=1$is in the interval $(-1,3)$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{x+1}{x鈭�3}=\frac{1+1}{1鈭�3}=\frac{2}{-2}=-1$

The quotient is marked negative in the interval $(-1,3)$.

The number $4$is in the interval $(3,鈭�)$. Test $x=4$in the expression in the numerator and the denominator.

$\frac{x+1}{x鈭�3}=\frac{4+1}{4鈭�3}=\frac{5}{1}=5$

The quotient is marked positive in the interval $(3,鈭�)$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(-鈭�,-1)$and $(3,鈭�)$are solutions.

The solution in interval notation is $(-鈭�,-1)鈭�(3,鈭�)$.