Question

$f(x) = \sqrt{|x|^2 - 5|x| + 6} + \sqrt{8 + 2|x| - |x|^2}$ is real for all x in

• A. [-4, -3]
• B. [-3, -2]
• C. [-2, 2]
• D. [3, 4]

Answer 1

Answer: (C)

[-2, 2]

Answer 2

To find the domain of the function $f(x) = \sqrt{|x|^2 - 5|x| + 6} + \sqrt{8 + 2|x| - |x|^2}$, we need to ensure that both terms under the square root are non-negative.

Let $y = |x|$. Since $|x| \ge 0$, we must have $y \ge 0$.

The first condition is for the term $\sqrt{|x|^2 - 5|x| + 6}$: $y^2 - 5y + 6 \ge 0$ Factor the quadratic: $(y-2)(y-3) \ge 0$ This inequality holds when $y \le 2$ or $y \ge 3$. Considering $y \ge 0$, this means $y \in [0, 2] \cup [3, \infty)$. (Condition 1)

The second condition is for the term $\sqrt{8 + 2|x| - |x|^2}$: $8 + 2y - y^2 \ge 0$ Rearrange the inequality: $y^2 - 2y - 8 \le 0$ Factor the quadratic: $(y-4)(y+2) \le 0$ This inequality holds when $-2 \le y \le 4$. (Condition 2)

Now, we need to find the intersection of Condition 1 and Condition 2, along with $y \ge 0$. The intersection of $y \in [0, 2] \cup [3, \infty)$ and $y \in [-2, 4]$ (which implicitly includes $y \ge 0$ for the relevant part of the intersection) is: $([0, 2] \cup [3, \infty)) \cap [-2, 4]$ $= ([0, 2] \cap [-2, 4]) \cup ([3, \infty) \cap [-2, 4])$ $= [0, 2] \cup [3, 4]$

So, the possible values for $y = |x|$ are $y \in [0, 2]$ or $y \in [3, 4]$.

Case 1: $0 \le |x| \le 2$ This implies $-2 \le x \le 2$. So, $x \in [-2, 2]$.

Case 2: $3 \le |x| \le 4$ This implies two possibilities: a) $3 \le x \le 4$. So, $x \in [3, 4]$. b) $3 \le -x \le 4 \implies -4 \le x \le -3$. So, $x \in [-4, -3]$.

Combining all possible ranges for $x$, the domain of $f(x)$ is $x \in [-4, -3] \cup [-2, 2] \cup [3, 4]$.

Since the question asks "is real for all x in" and multiple options (A, C, D) are correct subsets of the domain, and assuming it's a single-choice question, the most common convention for functions involving $|x|$ is to choose the interval symmetric about zero, which is option C.