Найдите область определения функции
y=[tex] \sqrt\frac{7-x} \sqrt{ 4x^{2}-19x+12 } [/tex]
y=[tex] \sqrt\frac{7-x} \sqrt{ 4x^{2}-19x+12 } [/tex]
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Verified answer
4x² - 19x + 12 > 0D = 361 - 192 = 169
x1 = (19-13)/8 = 6/8 = 3/4
x2 = (19+13)/8 = 32/8 = 4
(-∞; 3/4)U(4;+∞)
x - 7 ≥ 0
x ≥ 7
Итоговый ответ: [7; +∞)