Question

Знайти область визначення функції

Answer 1

$y = \sqrt{ {x}^{2} + 8x + 15 } \\ {x}^{2} + 8x +1 5 \geqslant 0 \\ \\ {x}^{2} + 8x + 5 = 0 \\ a =1 \\ b = 8 \\ c =15 \\ D = {b}^{2} - 4ac = {8}^{2} - 4 \times 1 \times 15 = 64 - 60 = 4 \\ x_{1} = \frac{ - 8 - 2}{2} = - \frac{10}{2} = - 5 \\ x_{2} = \frac{ -8 + 2}{2} = - \frac{6}{2} = - 3 \\ \\ (x + 5)(x + 3) \geqslant 0 \\ + + + [ - 5] - - - [ - 3] + + + \\ x \: \epsilon \: ( - \infty ; \: - 5]U[ - 3; \: + \infty )$