The domain of f(x)=x2−4|x|+3 is

f(x)=x2−4|x|+3=(|x|−3)(|x|−1)(|x|−3)(|x|−1)≥0|x|∈(−∞,1]∪[3,∞)|x|≤1 or |x|≥3x∈[−1,1]∪(−∞,−3]∪[3,∞)

The domain of f(x)=x2−4|x|+3 is

f(x)=x2−4|x|+3=(|x|−3)(|x|−1)(|x|−3)(|x|−1)≥0|x|∈(−∞,1]∪[3,∞)|x|≤1 or |x|≥3x∈[−1,1]∪(−∞,−3]∪[3,∞)

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The domain of $f (x) = \sqrt{x^{2} - 4 | x | + 3}$ is

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$(- \infty, 3]$

$[1, \infty)$

$(- \infty, - 3] \cup [- 1, \infty)$

$(- \infty, - 3] \cup [- 1, 1] \cup [3, \infty)$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{x^{2} - 4 | x | + 3} = \sqrt{(| x | - 3) (| x | - 1)} \\ (| x | - 3) (| x | - 1) \geq 0 \\ | x | \in (- \infty, 1] \cup [3, \infty) \\ | x | \leq 1 or | x | \geq 3 \\ x \in [- 1, 1] \cup (- \infty, - 3] \cup [3, \infty) \end{array}$