Question

The greatest value of $f\left(x\right)=(x+1{)}^{\frac{1}{3}}-(x-1{)}^{\frac{1}{3}}$ on $[0,1]$ is

• A. $1$
• B. $2$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$2$

$f\left(x\right)={(x+1)}^{1/3}-{(x-1)}^{1/3}$

$f^{\prime }\left(x\right)=\frac{1}{3}{(x+1)}^{-2/3}-\frac{1}{3}{(x-1)}^{-2/3}$ [for critical points]

$f^{\prime }\left(x\right)=0$

${(x+1)}^{-2/3}={(x-1)}^{-2/3}$

$\left(\frac{x+1}{x-1}\right)=1$

Thus no points could be found

Let $x=0$

$f\left(x\right)={\left(1\right)}^{1/3}-{(-1)}^{1/3}=1+1=2$

and at $x=1$

$f\left(x\right)={\left(2\right)}^{1/3}-0={\left(2\right)}^{1/3}$

Thus, greatest value of $f\left(x\right)$ is $2$