Answer
The inequality is valid for values only between -8 and -2 (including -8, but not -2 since it is restricted by the denominator) i.e. $ [-8,-2)$.
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x-4}{2x+4}\geq1$
$\dfrac{x-4}{2x+4}-1\geq0$
$\dfrac{x-4}{2x+4}-\dfrac{2x+4}{2x+4}\geq0$
$\dfrac{x-4-(2x+4)}{2x+4}\geq0$
$\dfrac{-x-8}{2x+4}\geq0$
Now, we find critical points by equating the numerator and denominator to zero:
$-x-8=0$
$2x+4=0$
There are two critical points:
$-x_1-8=0\rightarrow x_1=-8$
$2x_2+4=0\rightarrow x_2=-4/2=-2$
Next, we are going to take three values: one less than -8; one between -8 and -2; and one more than -2 to test in the original equation and check if the inequality is true or not:
First test with a value less than -8:
$\dfrac{-10-4}{2(-10)+4}\geq1$
$\dfrac{-14}{-20+4}\geq1$
$\dfrac{-14}{-16}\geq1$
$\dfrac{7}{8}\geq1 \rightarrow \text{ FALSE}$
Second test with a value between -8 and -2:
$\dfrac{-3-4}{2(-3)+4}\geq1$
$\dfrac{-7}{-6+4}\geq1$ $\dfrac{-7}{-2}\geq1$
$3.5\geq1 \rightarrow \text{ TRUE}$
Third test with a value more than -2:
$\dfrac{0-4}{2(0)+4}\geq1$
$\dfrac{-4}{4}\geq1$
$-1\geq1 \rightarrow \text{ FALSE}$
These tests show that the inequality $\dfrac{x-4}{2x+4}\geq1$ is valid for values only between -8 and -2 (including -8, but not -2 since it is restricted by the denominator) i.e. $ [-8,-2)$
