Question

The set of values for which $x^3+1\ge x^2+x$ is

• A. $x\ge 0$
• B. $x\le 0$
• C. $x\ge -1$
• D. $-1\le x\le 1$

Answer 1

The correct option is C $x\ge -1$

$x^3+1\ge x^2+x{x}^3-x^2-x+1\ge 0x^2(x-1)-1(x-1)\ge 0(x^2-1)(x-1)\ge 0(x-1{)}^2(x+1)\ge 0\therefore x\in [-1,\infty ]\therefore x\ge -1$