Question

Let $f\left(x\right)={\int }_0^1|t-x|tdt$ for all real x. Then the minimum value of f(x) is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}(1+\frac{1}{\sqrt{2}})$
• C. $\frac{1}{3}(1-\frac{1}{\sqrt{2}})$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{1}{3}(1-\frac{1}{\sqrt{2}})$

$f\left(x\right)=\{\begin{array}{c}\frac{1}{3}-\frac{x}{2}ifx\le 0\\ \frac{1}{3}+\frac{x^3}{3}-\frac{x}{2}if0\le x\le 1\\ \frac{x}{2}-\frac{1}{3}ifx\ge 1\end{array}$
f attains its minimum at $x=\frac{1}{\sqrt{2}}$