Answer
$[-2, -1) \cup [9, \infty)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{x}{2}\geq \frac{5}{x+1} + 4$,
$\displaystyle\frac{x}{2} -\frac{5}{x+1} -4\geq 0$ ,
$\displaystyle\frac{x^2+x-10-8x-8}{2(x+1)}\geq 0$ ,
$\displaystyle \frac{(x-9)(x+2)}{2(x+1)}\geq 0$ ,
$f(x)=\displaystyle \frac{(x-9)(x+2)}{2(x+1)}\geq 0$,
2.The cut points are:
$\displaystyle \frac{(x-9)(x+2)}{2(x+1)}=0$
$x=-2$ or $x=-1$ or $x=9$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& & \displaystyle \displaystyle \frac{(a-9)(a+2)}{2(a+1)} & \\
(-\infty,-2) & -5 & \frac{(-)(-)}{(-)}=(-) & F\\
(-2, -1) & -\frac{3}{2} & \frac{(-)(+)}{(-)}=(+) & T\\
(-1, 9) & 0 & \frac{(-)(+)}{(+)}=(-) & F\\
(9,\infty) & 10 & \frac{(+)(+)}{(+)}=(+) & T\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points unless they are zeros of the denominator.
Solution set: $[-2, -1) \cup [9, \infty)$