Question

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

• A. 1
• B. 2
• C. 3
• D. 1/3

Answer 1

Answer: (B)

We have $f(x)={(x+1)}^{1/3}-{(x-1)}^{1/3}$ $\therefore$ $f(x)=\frac{1}{3}\left[\frac{1}{{(x+1)}^{2/3}}-\frac{1}{{(x-1)}^{2/3}}\right]$ $=\frac{{(x-1)}^{2/3}-{(x+1)}^{2/3}}{3{(x^2-1)}^{2/3}}$ Clearly, $f(x)$ does not exists at $x=\pm 1$ Now, $f(x)\ne 0$ then ${(x-1)}^{2/3}={(x+1)}^{2/3}$ $\Rightarrow$ $x=0$ Clearly $f(x)\ne 0$ for any other value of $x\in [0,1]$ . The value of $f(x)$ at $x=0$ is 2. Hence, the greatest value of $f(x)$ is 2