Question

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

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

Answer 1

The correct option is C maximum value of $f$ is $9$

$f\left(x\right)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}}$
$\Rightarrow f^{\prime }\left(x\right)=8x^{\frac{1}{3}}-x^{-\frac{2}{3}}$

For criical points, $f^{\prime }\left(x\right)=0$
$\Rightarrow 8x^{\frac{1}{3}}-x^{-\frac{2}{3}}=0$
$\Rightarrow x=\frac{1}{8}$

$\Rightarrow f\left(x\right){|}_{x=\frac{1}{8}}=-\frac{9}{8}$
$\Rightarrow f\left(x\right){|}_{x=1}=6-3=3$
$\Rightarrow f\left(x\right){|}_{x=-1}=6+3=9$
Maximum value of $f\left(x\right)=9$