Question

Find the value of $x$ for which $f\left(x\right)=(1-2x{)}^2$ will attain its minimum value.

• A. $-1$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{2}$
• E. $1$

Answer 1

The correct option is D $\frac{1}{2}$

Given, $f\left(x\right)=(1-2x{)}^2$

The minimum value of any square term is $0$.
Therefore, the minimum value of $f\left(x\right)$ will occur when $1-2x=0$, which gives $x=\frac{1}{2}$.