Question

Find values of x simultaneously satisfying: $\begin{cases} x^2 - 4 \geq 0 \\ x^2 - 2x - 8 \geq 0 \\ -x^2 + 5x - 4 \geq 0 \end{cases}$

Answer 1

x = 4

Answer 2

Solving the system of inequalities involves finding the intersection of the solution sets for each inequality.

1. Solve $x^2 - 4 \geq 0$. This gives $x \in (-\infty, -2] \cup [2, \infty)$.

2. Solve $x^2 - 2x - 8 \geq 0$. This gives $x \in (-\infty, -2] \cup [4, \infty)$.

3. Solve $-x^2 + 5x - 4 \geq 0$, which is equivalent to $x^2 - 5x + 4 \leq 0$. This gives $x \in [1, 4]$.

The intersection of these three solution sets is the set of values that satisfy all three inequalities. Taking the intersection of the first two solution sets gives $(-\infty, -2] \cup [4, \infty)$. Then, intersecting this with the third solution set $[1, 4]$ yields $\{4\}$. Therefore, the only value of $x$ that satisfies all three inequalities is $x = 4$.