Question

знайдіть область визначення функції​

Answer 1

$\displaystyle\bf\\f(x)=\frac{\sqrt{14+5x-x^{2} } }{x^{2} +x-6} \\\\\\\left \{ {{14+5x-x^{2} \geq 0} \atop {x^{2} +x-6\neq 0}} \right. \\\\\\1)\\\\14+5x-x^{2} \geq 0\\\\x^{2} -5x-14\leq 0\\\\(x^{2} -4)-(5x+10)\leq 0\\\\(x+2)(x-2)-5(x+2)\leq 0\\\\\\(x+2)\cdot(x-2-5)\leq 0\\\\(x+2)\cdot(x-7)\leq 0\\\\\\+ + + + + \Big[-2\Big] - - - - - \Big[7 \Big] + + + + + \\\\\\\boxed{x\in\Big[-2 \ ; \ 7\Big]}\\\\2)\\\\x^{2} +x-6\neq 0\\\\(x-2)\cdot(x+3)\neq 0$

$\displaystyle\bf\\x-2\neq 0 \ \ \ \Rightarrow \ \ \ x\neq 2\\\\x+3\neq 0 \ \ \ \Rightarrow \ \ \ x\neq -3\\\\\\Otvet \ : \ D(f)=\Big[-2 \ ; \ 2\Big)\cup\Big(2 \ ; \ 7\Big]$

9 целых чисел  :  -2 ; -1 ; 0 ; 1 ; 3 ; 4 ; 5 ; 6 ; 7