Question

The maximum value of ${[x(x-1)+1]}^{\frac{1}{3}}$, $0\le x\le 1$ is

• A. ${\left(\frac{3}{4}\right)}^{\frac{1}{3}}$
• B. $\frac{1}{2}$
• C. $1$
• D. $0$

Answer 1

The correct option is A ${\left(\frac{3}{4}\right)}^{\frac{1}{3}}$

$f\left(x\right)={\left[x\right(x-1)+1]}^{\frac{1}{3}}0\le x\le 1$

Now, $f\left(x\right)=0$ (for critical points)

$f^{\prime }\left(x\right)=\frac{1}{3}{\left[x\right(x-1)+1]}^{\frac{1}{3}-1}\times [2x-1]=0$

Now, $x=\frac{1}{2}$ is the only point between $0$ and $1$

$f\left(\frac{1}{2}\right)={[\frac{1}{2}(\frac{1}{2}-1)+1]}^{\frac{1}{3}}={[-\frac{1}{4}+1]}^{\frac{1}{3}}$

$f\left(\frac{1}{2}\right)={\left(\frac{3}{4}\right)}^{\frac{1}{3}}$ (Maximum value)