Question

Find the greatest & the least value of the following function in the given interval, if it exist.

$f\left(x\right)=12x^{4/3}-6x^{1/3},xϵ[-1,1]$

• A. Maximum is 18 and minimum is 5
• B. Maximum is 7 and minimum is 0
• C. Maximum is 7 and minimum is -9/4
• D. Maximum is 18 and minimum is -9/4

Answer 1

The correct option is D Maximum is 18 and minimum is -9/4

Consider, $f\left(x\right)=12x^{4/3}-6x^{1/3}x\in [-1,1]$

$f^{\prime }\left(x\right)=16x^{1/3}-2x^{-2/3}$

$f^{\prime }\left(x\right)>0\Rightarrow 16x^{1/3}>2x^{-2/3}\Rightarrow 16x>2\Rightarrow x>\frac{1}{8}$

$f^{\prime }\left(x\right)<0\Rightarrow 16x^{1/3}<2x^{-2/3}\Rightarrow x<\frac{1}{8}$

$\therefore f^{\prime }\left(x\right)$ increases on $(\frac{1}{8},1)$ and decreases on $(-1,\frac{1}{8})$

Also $f\left(1\right)=6$

$f(-1)=18$

$f\left(\frac{1}{8}\right)=-\frac{9}{4}$

$\therefore$ Min value $=-\frac{9}{4}$ and max value $=18$