Question

Найдите область определения функции:
y = (корень)
16-х^2/x-3 (без корня)

Answer 1

$y= \frac{\sqrt{16-x^2}}{x-3}\\\\OOF: \left \{ {{16-x^2 \geq 0} \atop {x-3\ne 0}} \right. \geq ; \left \{ {{x^2-16 \leq 0} \atop {x\ne 3}} \right. \; \left \{ {{(x-4)(x+4) \leq 0} \atop {x\ne 3}} \right. \; \left \{ {{x\in [-4,4\, ]} \atop {x\ne 3}} \right. \\\\x\in [-4,3)\cup (3,4\, ]$