Question

If $x$ is a natural number , then the minimum value of$(x^2-6x+12)$ is

• A. 3
• B. 7
• C. 6
• D. 12

Answer 1

The correct option is A 3

Given: $f\left(x\right)=x^2-6x+12$

Minimum value of $f\left(x\right)$ is at $x=x_1$, where $x_1$ is that value of $x$ where $f^{\prime }\left(x\right)=0$

Here,

$f^{\prime }\left(x\right)=2x-6$

So,

$f^{\prime }\left(x\right)=0=2x-6$

Hence, $x=x_1=3$

Therefore maximum value of $f\left(x\right)$ is

$=f\left(3\right)=3^2-6\left(3\right)+12=21-18=3$