Question

Let $f\left(x\right)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}$ defined on $[-1,1]$ then

• A. maximum value of $f\left(x\right)$ is $7$
• B. maximum value of $f\left(x\right)$ is $5$
• C. maximum value of $f\left(x\right)$ is $9$
• D. maximum value of $f\left(x\right)$ is $-\frac{3}{2}$

Answer 1

Answer: (C)

maximum value of $f\left(x\right)$ is $9$

$f\left(x\right)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}$
$\Rightarrow f^{\prime }\left(x\right)=6\times \frac{4}{3}{x}^{\frac{1}{3}}-3\times \frac{1}{3}{x}^{\frac{-2}{3}}$
$\Rightarrow f^{\prime }\left(x\right)=8x^{\frac{1}{3}}-x^{\frac{-2}{3}}$
For maximum/minimum, we must have
$f^{\prime }\left(x\right)=0$
$\Rightarrow 8x^{\frac{1}{3}}-x^{\frac{-2}{3}}=0$
$\Rightarrow 8x-1=0$
$\Rightarrow x=\frac{1}{8}$
Now, $f(-1)=6+3=9$
$f\left(\frac{1}{8}\right)=\frac{3}{8}-\frac{3}{2}=-\frac{9}{8}$
$f\left(1\right)=6-3=3$
Option C is correct.