Question

Find the absolute maximum and minimum values of a function f given by

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

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}} \\ \Rightarrow f\text{'}\left(x\right)=16x^{\frac{1}{3}}-2x^{\frac{-2}{3}}=\frac{2\left(8x-1\right)}{x^{{\displaystyle \frac{2}{3}}}} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{2\left(8x-1\right)}{x^{{\displaystyle \frac{2}{3}}}}=0 \\ \Rightarrow 8x-1=0 \\ \Rightarrow x=\frac{1}{8} \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}\mathit{}f\mathrm{are}-1,\frac{1}{8}\mathrm{and}1. \\ \mathrm{Now}, \\ f\left(-1\right)=12{\left(-1\right)}^{\frac{4}{3}}-6{\left(-1\right)}^{\frac{1}{3}}=18 \\ f\left(\frac{1}{8}\right)=12{\left(\frac{1}{8}\right)}^{\frac{4}{3}}-6{\left(\frac{1}{8}\right)}^{\frac{1}{3}}=\frac{-9}{4} \\ f\left(1\right)=12{\left(1\right)}^{\frac{4}{3}}-6{\left(1\right)}^{\frac{1}{3}}=6 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=-1\mathrm{is}18\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}x=\frac{1}{8}\mathrm{is}\frac{-9}{4}. \end{gathered}$$