Questions
- For each rational function \(f\), use the asymptotic behavior and its behavior near its zeros and poles to sketch \(f.\)
- \(f(x)=\frac{3x-1}{x^2+x}\)
- \(f(x)=\frac{(x+4)(x-5)^3(x-9)^2}{5x^2(x^2-1)(x+5)^2}\)
- \(f(x)=\frac{5x+4}{x^2+6}\)
- \(f(x)=\frac{x^4+1}{x+1}\)
- For each polynomial \(f\), solve each inequality separately: \(f(x)\geq 0\), \(f(x)>0\), \(f(x)<0\) and \(f(x)\leq 0\).
- \(f(x)=2(x+3)^2(x+2)^3(x-5)^3(x-6)^8\)
- \(f(x)=5x^3(x+5)(x+1)^6(2x-3)^6(x-4)^7\)
- \(f(x)=4(x+5)^4(5x+2)^3(5-2x)^5(x-6)\)
- \(f(x)=-5x(x+5)^4(5x+2)^2(2x-5)(x-6)^3\)
- For each rational function \(f\), solve each inequality separately: \(f(x)\geq 0\), \(f(x)>0\), \(f(x)<0\) and \(f(x)\leq 0\).
- \(f(x)=\frac{1}{x^2(x-4)}\)
- \(f(x)=\frac{1}{-5(x+3)(x+1)(x-5)^2}\)
- \(f(x)=\frac{1}{-3x(x+3)(x+1)^2(x-3)}\)
- \(f(x)=\frac{3x-1}{x^2+x}\)
- \(f(x)=\frac{(x+4)(x-5)^3(x-9)^2}{5x^2(x^2-1)(x+5)^2}\)
- \(f(x)=\frac{5x+4}{x^2+6}\)
- \(f(x)=\frac{x^4+1}{x+1}\)
Answers
- solution to \(f(x)\geq 0\) is \((-\infty,-2]\cup[5,\infty)\);
solution to \(f(x)>0\) is \((-\infty,-3)\cup(-3,-2)\cup(5,6)\cup(6,\infty)\);
solution to \(f(x)<0\) is \((-2,5)\);
solution to \(f(x)\leq 0\) is \(\{-3\}\cup[-2,5]\cup\{6\}.\)
- solution to \(f(x)\geq 0\) is \([-5,0]\cup\{1.5\}\cup[4,\infty)\);
solution to \(f(x)>0\) is \((-5,-1)\cup(-1,0)\cup(4,\infty)\);
solution to \(f(x)<0\) is \((-\infty,-5)\cup(0,-1.5)\cup(-1.5,4)\);
solution to \(f(x)\leq 0\) is \((-\infty,-5]\cup\{-1\}\cup[0,4].\)
- solution to \(f(x)\geq 0\) is \(\{-5\}\cup[-0.4,2.5]\cup[6,\infty)\);
solution to \(f(x)>0\) is \((-0.4,2.5)\cup(6,\infty)\);
solution to \(f(x)<0\) is \((-\infty,-5)\cup(2.5,6)\);
solution to \(f(x)\leq 0\) is \((-\infty,-5]\cup\{-0.4\}\cup[2.5,6].\)
- solution to \(f(x)\geq 0\) is \((-\infty,0]\cup[2.5,6]\);
solution to \(f(x)>0\) is \((-\infty,-5)\cup(-5,-0.4)\cup(-0.4,0)\cup(2.5,6)\);
solution to \(f(x)<0\) is \((0,2.5)\cup(6,\infty)\);
solution to \(f(x)\leq 0\) is \(\{-5\}\cup\{-0.4\}\cup[0,2.5]\cup[6,\infty).\)
- For each rational function \(f\), solve each inequality separately: \(f(x)\geq 0\), \(f(x)>0\), \(f(x)<0\) and \(f(x)\leq 0\).
- solution to \(f(x)\geq 0\) is \((4,\infty)\);
solution to \(f(x)>0\) is \((4,\infty)\);
solution to \(f(x)<0\) is \((-\infty,0)\cup(0,4)\);
solution to \(f(x)\leq 0\) is \((-\infty,0)\cup(0,4).\)
- solution to \(f(x)\geq 0\) is \((-3,-1)\);
solution to \(f(x)>0\) is \((-3,-1)\);
solution to \(f(x)<0\) is \((-\infty,-3)\cup(-1,5)\cup(5,\infty)\);
solution to \(f(x)\leq 0\) is \((-\infty,-3)\cup(-1,5)\cup(5,\infty).\)
- solution to \(f(x)\geq 0\) is \((-3,-1)\cup(-1,0)\cup(3,\infty)\);
solution to \(f(x)>0\) is \((-3,-1)\cup(-1,0)\cup(3,\infty)\);
solution to \(f(x)<0\) is \((-\infty,-3)\cup(0,3)\);
solution to \(f(x)\leq 0\) is \((-\infty,-3)\cup(0,3).\)
- solution to \(f(x)\geq 0\) is \((-1,0)\cup\left[\frac{1}{3},\infty\right)\);
solution to \(f(x)>0\) is \((-1,0)\cup\left(\frac{1}{3},\infty\right)\);
solution to \(f(x)<0\) is \((-\infty,-1)\cup\left(0,\frac{1}{3}\right)\);
solution to \(f(x)\leq 0\) is \((-\infty,-1)\cup\left(0,\frac{1}{3}\right].\)
- solution to \(f(x)\geq 0\) is \((-\infty,-5)\cup(-5,-4]\cup(-1,0)\cup(0,1)\cup[5,\infty)\);
solution to \(f(x)>0\) is \((-\infty,-5)\cup(-5,-4)\cup(-1,0)\cup(0,1)\cup(5,9)\cup(9,\infty)\);
solution to \(f(x)<0\) is \((-4,-1)\cup(1,5)\);
solution to \(f(x)\leq 0\) is \([-4,-1)\cup(1,5].\)
- solution to \(f(x)\geq 0\) is \(\left[-\frac{4}{5},\infty\right)\);
solution to \(f(x)>0\) is \(\left(-\frac{4}{5},\infty\right)\);
solution to \(f(x)<0\) is \(\left(-\infty,-\frac{4}{5}\right)\);
solution to \(f(x)\leq 0\) is \(\left(-\infty,-\frac{4}{5}\right].\)
- solution to \(f(x)\geq 0\) is \((-1,\infty)\);
solution to \(f(x)>0\) is \((-1,\infty)\);
solution to \(f(x)<0\) is \((-\infty,-1)\);
solution to \(f(x)\leq 0\) is \((-\infty,-1).\)