Question

Find the points of absolute maximum and minimum of $f\left(x\right)=(x-1{)}^{1/3}(x-2);1\le x\le 9$.

Answer 1

$f\left(x\right)={(x-1)}^{1/3}(x-2);1\le x\le 9f^{\prime }\left(x\right)=\frac{1}{3}{(x-1)}^{-2/3}(x-2)+{(x-1)}^{1/3}=0=\frac{(x-2)}{3{(x-1)}^{2/3}}+{(x-1)}^{1/3}=0\Rightarrow (x-2)+3(x-1)=0\Rightarrow 4x-5=0\Rightarrow x=\frac{5}{4}{f}^{\prime \prime }\left(x\right)=\frac{1}{3}\left(\frac{-2}{3}\right){(x-1)}^{-5/3}(x-2)+\frac{1}{3}{(x-1)}^{-2/3}+\frac{1}{3}{(x-1)}^{-2/3}=\frac{2}{3}{(x-1)}^{-2/3}\left[\frac{1}{3}{(x-1)}^{-1}\right(x-2)+1]=\frac{2}{3}{(x-1)}^{-2/3}[\frac{(x-2){(x-1)}^{-1}}{3}+1]f^{\prime \prime }\left(x\right)>0forx=\frac{5}{4}$

Therefore, $x=\frac{5}{4}$ is the absolute minima

Absolute minima will then be at $x=9$