Question

What is the value of $x$, when $f\left(x\right)=6+(x-2{)}^2$ is at its minimum?

• A. $-6$
• B. $-4$
• C. $0$
• D. $2$
• E. $5$

Answer 1

The correct option is D $2$

Given, $f\left(x\right)=6+{(x-2)}^2$
For $f\left(x\right)$ to be minimum, $6+{(x-2)}^2$ must be minimum, which implies ${(x-2)}^2$ should be minimum.
The minimum value of any square term is $0$.
So, the minimum value of ${(x-2)}^2$ is $0$ and occurs at $x=2$.