The function  $f\left(x\right)=x^{1/3}(x-1)$

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{1/3}\cdot (\mathrm{x}-1)\Rightarrow \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{4}{3}{\mathrm{x}}^{1/3}-\frac{1}{3}\cdot \frac{2}{{\mathrm{x}}^{2/3}}=\frac{1}{3{\mathrm{x}}^{2/3}}[4\mathrm{x}-1]$
$\Rightarrow$ f(x) changes sign from – ve to +ve, at x = ¼, which is point of minima. Also does not exist at x = 0 as f(x) has vertical tangent at x=0
${\mathrm{f}}^{11}\left(\mathrm{x}\right)=\frac{4}{9}\cdot \frac{1}{{\mathrm{x}}^{2/3}}+\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{1}{{\mathrm{x}}^{5/3}}\Rightarrow \frac{2}{9{\mathrm{x}}^{2/3}}\left[2+\frac{1}{\mathrm{x}}\right]=\frac{2}{9{\mathrm{x}}^{2/3}}\left[\frac{2\mathrm{x}+1}{\mathrm{x}}\right]$
${\text{∴f}}^{11}\left(\mathrm{x}\right)=0 \mathrm{at} \mathrm{x}=-\frac{1}{2}$ which is the point of inflection at x=0,  f¹(x) doesn’t exists  f¹¹(x) changes sign, hence x = 0 is also the point of inflection. From the above information the graph of y = f(x) is as shown

Also, minimum value of f(x) is t x = 1/4 which is -3 x 2^-8/3  Hence, range is [-3 x 2^-8/3, $\infty$)