Question

The greatest value of $y={(x+1)}^{1/3}-{(x-1)}^{1/3}$ on $[0,1]$ is

• A. $1$
• B. $2$
• C. $3$
• D. $2^{1/3}$

Answer 1

The correct option is B $2$

$f\left(x\right)=(x+1{)}^{\frac{1}{3}}-(x-1{)}^{\frac{1}{3}}$
$f^{\prime }\left(x\right)$
$=\frac{1}{3}\left[\right(x+1{)}^{\frac{-2}{3}}-(x-1{)}^{\frac{-2}{3}}]$
$=0$
Or
$(x+1{)}^{\frac{-2}{3}}=(x-1{)}^{\frac{-2}{3}}$
Or
$(x-1{)}^{\frac{2}{3}}=(x+1{)}^{\frac{2}{3}}$
Or
${\frac{(x+1)}{(x-1)}}^{\frac{2}{3}}=1$
Or
$\frac{x+1}{x-1}=1^{\frac{3}{2}}$
Or
$\frac{x+1}{x-1}=\pm 1$
Now
Considering
$x+1=-(x-1)$
Or
$2x=0$
Or
$x=0$
$f"\left(x\right)$
$=\frac{-2}{9}\left[\right(x+1{)}^{\frac{-5}{3}}-(x-1{)}^{\frac{-5}{3}}]$
$f"\left(0\right)<0$
Hence it attains a maxima at $x=0$
Now
$f\left(0\right)=1-(-1{)}^{\frac{1}{3}}$
$=1-(-1)$
$=2$.
Hence maximum value is 2.