Given: `g(x)=x^(1/3)-x^(-2/3)`

Find the critical value(s) of the function by setting the first derivative equal to zero and solving for the x value(s).

`g'(x)=(1/3)x^(-2/3)+(2/3)x^(-5/3)=0`

`1/(3x^(2/3))=-2/(3x^(5/3))`

`3x^(5/3)=-6x^(2/3)`

`3x^(5/3)+6x^(2/3)=0`

`3x^(2/3)(x+2)=0`

`x=0,x=-2`

The value x=0 is not defined in g(x).

The critical value is x=-2.

Convert to fractional form :

`g(x) = x^(1/3) - 1/x^(2/3) `

`g(x) = (x - 1)/x^(2/3) `

Vertical asymptotes occur when denominator = 0.

If `x^(2/3) = 0` , then `x = 0` , so vertical asymptote at `x = 0` .

Take the derivative and set the numerator equal to 0.

`g'(x) = (x + 2)/[3x^(5/3)] = 0`

`x = -2 `

Substituting `x = -2` into g(x) gives : `g(-2) = -1.89` approx.

Therefore `(-2, -1.89)` is a critical point.