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The function f(x)=x^(1/3)(x-1) has two ...

The function `f(x)=x^(1/3)(x-1)` has two inflection points has one point of extremum is non-differentiable has range `[-3x2^(-8/3),oo)`

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`y=x^(1//3)(x-1)`
`(dy)/(dx)=4/3x^(1//3)-1/3.1/x^(2//3) =1/(3x^(2//3)) [4x-1]`
Hence f is increasing for `x gt 1/4` and decreasing for `x lt 1/4`
Thus, `x=1//4` is the point of minima.
`f^(')(0)` does not exist as `f(x)` is a vertical tangent here.
`f^('')(x) = 4/9. 1/x^(2//3) + 1/3. 2/3. 1/x^(5//3)`
`=2/(9x^(2//3))[2+1/x]=2/(9x^(2//3))[(2x+1)/(x)]`
`therefore f^('')(x)=0` at `x=-1/2`
Thus, `x=-1/2` is the point of inflection.
From the above discussion, the graph of function is as shown in the following figure.
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