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 $x\ge -1$

Best way to solve this problem is by checking different options.
For x = -1,
$x^3+1=0and{x}^2+x=0$,
hence, $x^3+1=x^2+x=0$
So, option (a) is eliminated.
For x = 2,
$x^3+1=9and{x}^2+x=6$,
hence, $x^3+1\ge x^2+x$ for x = 2.
Hence, option (b) and (d) are also eliminated.
$\therefore$ Option (c) is correct answer.

Alternate Approach:
$x^3+1\ge x^2+x\Rightarrow x^3-x^2-x+1\ge 0\Rightarrow (x-1{)}^2(x+1)\ge 0\Rightarrow x\ge -1$