Question

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

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

Answer 1

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

Given, $\left[x\right(x-1)+1{]}^{\frac{1}{3}}$

Derivative of this function will be

$\frac{1}{3}{\left[x\right(x-1)-1]}^{\frac{-2}{3}}(2x-1)$

Now let's put this equal to zero, so we get $x=\frac{1}{2}$.

Now you can find double derivative of this function and see it's value at $x=\frac{1}{2}$ and you will see that it is negative.

So, function will take maxima at $x=\frac{1}{2}$ and the maximum value is ${\left(\frac{3}{4}\right)}^{\frac{1}{3}}$.