The critical points of the function $f(x)={ x }^{ 3/5 }\left( 4-x \right) ,x\in { R }^{ + }\cup \left\{ 0 \right\}$ is _______

$f(x)={ x }^{ \cfrac { 3 }{ 5 } }\left( 4-x \right) x\in { R }^{ + }$

$f(x)=4{ x }^{ \cfrac { 3 }{ 5 } }-{ x }^{ \cfrac { 8 }{ 5 } }$

$\therefore f'(x)=4\left( \cfrac { 3 }{ 5 } \right) { x }^{ -\cfrac { 2 }{ 3 } }=\cfrac { 8 }{ 5 } { x }^{ \cfrac { 3 }{ 5 } }$

$\therefore f'(x)=\cfrac { 4 }{ 5 } \left( \cfrac { 3 }{ { x }^{ \cfrac { 2 }{ 3 } } } -2{ x }^{ \cfrac { 3 }{ 5 } } \right) =\cfrac { 4 }{ 5 } \left( \cfrac { 3-2x }{ { x }^{ \cfrac { 2 }{ 3 } } } \right) \quad$

For $x=0;f'(x)$ does not exists and for $x=\cfrac { 2 }{ 3 }$ we have $f'(x)=0$

$\therefore$ Critical points are $0$ and $\cfrac{2}{3}$