Digital ACT Math Practice Question #2116
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Test boundary $x = 3$: $\sqrt{0}/(3-5) = 0/-2 = 0$ (valid! must include bracket [3). Test $x = 5$: division by zero (undefined! must exclude 5).
For the square root in the numerator to be defined over real numbers, $x - 3 \ge 0 \implies x \ge 3$. For the denominator to be nonzero, $x - 5 \ne 0 \implies x \ne 5$. Combining both restrictions gives $[3, 5) \cup (5, \infty)$. Note that ...
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