Question

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

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

Answer 1

Answer: (C)

1

Answer 2

Let $ƒ(x)=[x(x-1)+1]\frac{1}{3}.$

$∴ƒ'(x)=\frac{2x-1}{3[x(x-1)+1]\frac{2}{3}}$

Now$,ƒ'(x)=0⇒x=\frac{1}{2}$

Then,we evaluate the value of f at critical point $\frac{1}{2}$ and at the end points of the

interval $[0,1]${i.e.,at $x=0 and x=1$}.

$ƒ(0)=[0(0-1)+1]\frac{1}{3}=1$

$ƒ(1)=[1(1-1)+1]\frac{1}{3}=1$

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

Hence,we can conclude that the maximum value of $f$ in the interval $[0,1]is 1.$

The correct answer is C.