Question

The set $\{x^3-12x:-3\le x\le 3\}$ is equal to

• A. $\{x:-16\le x\le 16\}$
• B. $\{x:-12\le x\le 12\}$
• C. $\{x:-9\le x\le 9\}$
• D. $\{x:0\le x\le 10\}$

Answer 1

Answer: (A)

Let $y=x^3-12x$
$\frac{dy}{dx}=3x^2-12$
Put $\frac{dy}{dx}=0,3x^2-12=0$
$\Rightarrow$ $x=\pm 2$
At $x=2,y=2^3-12(2)=-16$
At $x=-2,y={(-2)}^3-12(-2)=16$
Hence, option is correct.