Найдите промежутки знакопостоянства функции y=x^3-2x^2-8x/x^2-1 y=x^2-9/x^2-4 y=x^2+4/x^2+3x Хотя бы одно пожалуйста!)
Числовые промежутки, на которых функция сохраняет свой знак (т.е. остается положительной или отрицательной), называются промежутками знакопостоянства функции.
![1)\; \; y=\frac{x^3-2x^2-8x}{x^2-1} \\\\y=0\; \to \; \left \{ {{x^3-2x^2-8x=0} \atop {x^2-1\ne 0}} \right. \; \left \{ {{x(x^2-2x-8)=0} \atop {(x-1)(x+1)\ne 0}} \right. \; \left \{ {{x(x-4)(x+2)=0} \atop {x\ne -1,\; x\ne 1}} \right. \\\\---[-2\, ]+++(-1)---[\, 0\, ]+++(1)---[\, 4\, ]+++\\\\y\ \textgreater \ 0:\; \; x\in (-2,-1)\cupc(0,1)\cup (4,+\infty )\\\\y\ \textless \ 0:\; \; x\in (-\infty ,-2)\cup (-1,0)\cup (1,4)](https://tex.z-dn.net/?f=1%29%5C%3B+%5C%3B++y%3D%5Cfrac%7Bx%5E3-2x%5E2-8x%7D%7Bx%5E2-1%7D+%5C%5C%5C%5Cy%3D0%5C%3B+%5Cto+%5C%3B+%5Cleft+%5C%7B+%7B%7Bx%5E3-2x%5E2-8x%3D0%7D+%5Catop+%7Bx%5E2-1%5Cne+0%7D%7D+%5Cright.+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%28x%5E2-2x-8%29%3D0%7D+%5Catop+%7B%28x-1%29%28x%2B1%29%5Cne+0%7D%7D+%5Cright.+%5C%3B+++%5Cleft+%5C%7B+%7B%7Bx%28x-4%29%28x%2B2%29%3D0%7D+%5Catop+%7Bx%5Cne+-1%2C%5C%3B+x%5Cne+1%7D%7D+%5Cright.++%5C%5C%5C%5C---%5B-2%5C%2C+%5D%2B%2B%2B%28-1%29---%5B%5C%2C+0%5C%2C+%5D%2B%2B%2B%281%29---%5B%5C%2C+4%5C%2C+%5D%2B%2B%2B%5C%5C%5C%5Cy%5C+%5Ctextgreater+%5C+0%3A%5C%3B+%5C%3B+x%5Cin+%28-2%2C-1%29%5Ccupc%280%2C1%29%5Ccup+%284%2C%2B%5Cinfty+%29%5C%5C%5C%5Cy%5C+%5Ctextless+%5C+0%3A%5C%3B+%5C%3B+x%5Cin+%28-%5Cinfty+%2C-2%29%5Ccup+%28-1%2C0%29%5Ccup+%281%2C4%29 "1)\; \; y=\frac{x^3-2x^2-8x}{x^2-1} \\\\y=0\; \to \; \left \{ {{x^3-2x^2-8x=0} \atop {x^2-1\ne 0}} \right. \; \left \{ {{x(x^2-2x-8)=0} \atop {(x-1)(x+1)\ne 0}} \right. \; \left \{ {{x(x-4)(x+2)=0} \atop {x\ne -1,\; x\ne 1}} \right. \\\\---[-2\, ]+++(-1)---[\, 0\, ]+++(1)---[\, 4\, ]+++\\\\y\ \textgreater \ 0:\; \; x\in (-2,-1)\cupc(0,1)\cup (4,+\infty )\\\\y\ \textless \ 0:\; \; x\in (-\infty ,-2)\cup (-1,0)\cup (1,4)")
