Question

The greatest value of $\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{(\mathrm{x}-1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ on $\left[0,1\right]$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

The explanation for the correct answer.

Find the greatest value.

$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{(\mathrm{x}-1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Take the derivative of $\mathrm{f}\left(\mathrm{x}\right)$.

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{3}{(\mathrm{x}+1)}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-\frac{1}{3}{(\mathrm{x}-1)}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{3}\left(\frac{1}{{(\mathrm{x}+1)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{1}{{(\mathrm{x}-1)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\right) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{3}\left(\frac{{(\mathrm{x}-1)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-{(\mathrm{x}+1)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{{\left({\mathrm{x}}^2-1\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\right) \end{gathered}$$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)$does not exist at $\mathrm{x}=\pm 1$

Put $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$ then ${(\mathrm{x}-1)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={(\mathrm{x}+1)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{x}=0,\mathrm{f}\left(\mathrm{x}\right)={(0+1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{(0-1)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=2$

Hence, option $B$ is the correct answer.