Question

Find the value of $x$ for which $x{(1-x)}^2,0\le x\le 2.$ is max.

• A. 1
• B. 2
• C. 1/3
• D. 3

Answer 1

Answer: (C)

1/3

Given, $f\left(x\right)=x(1-x{)}^2$
Now, $f^{\prime }\left(x\right)=(1-x{)}^2-2x(1-x)=0$
$(1-x)[1-x-2x]=0$
$(1-x)[1-3x]=0$
$x=1$ and $x=\frac{1}{3}$.
$f"\left(x\right)$ $=-2(1-x)+2x-2(1-x)$
$=2[x-2(1-x\left)\right]$
$=2x-2+2x$
$=4x-2$
Now, $f"\left(\frac{1}{3}\right)$
$=\frac{4}{3}-2$
$=\frac{-2}{3}$.
Hence, $f"\left(\frac{1}{3}\right)<0$
Thus $f\left(x\right)$ attains a maximum at $x=\frac{1}{3}$