Question

Let $f\left(x\right)=6x^{4/3}-x^{1/3}$ 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. minimum value of $f$ is -3/2

Answer 1

The correct option is A maximum value of $f$ is 7

$f^{\prime }\left(x\right)$
$=8x^{\frac{1}{3}}-\frac{x^{\frac{-2}{3}}}{3}$
$=0$
Or
$8x^{\frac{1}{3}}=\frac{x^{\frac{-2}{3}}}{3}$
Or
$8x=\frac{1}{3}$
$x=\frac{1}{24}$.
$f\left(\frac{1}{24}\right)$
$=6.x.x^{\frac{1}{3}}-x^{\frac{1}{3}}$
$=\frac{{24}^{\frac{1}{3}}}{4}-{24}^{\frac{1}{3}}$
$=\frac{-3}{4}{.24}^{\frac{1}{3}}$..(i)
$f\left(1\right)=5$ ...(ii)
and
$f(-1)=7$...(iii)
Hence from i, ii, and iii, we get the maximum value of $f\left(x\right)$ is 7.