Discuss the extremum of `f(x)=x(x^2-4)^(-1/3)`

Answer

Point of local maxima : x=`-2sqrt(3)`
Point of local minima : `x=2sqrt(3)`

Explanation

`f(x)=x(x^(2)-4)^(-1//3`
`f(x)=(x^(2)-4)^(-1//3-1)/(3)(x^(2)-4)-4//3(2x)x`
`=(1)/(x^(2)-4)^(1//3)-(2x^(2))/(3(x^(2)-4)^(4//3)`
`=3(x^(2)-4)-2x^(2)/(3(x^(2-4))^(4//3)`
`=(x^(2-12))/(3(x^(2)-4)^(4//3)`
Sign scheme of f(X) is as follows:
Thus `x=2 sqrt(3)` is point opf minima and x =`-sqrt(3)` is point of maxima
`f_(min)=f(2 sqrt(3))=2 sqrt(3)(12-4)A^(-1//3)=sqrt(3)`
Here maxima value is less than minimum value
Thus is because f(X) discontinous x=2
Since f(X) is unbounded function both the extreme values are local