You need to determine the monotony of the function, hence, you need to find the intervals where f'(x)>0 or f'(x)<0.

You need to find the derivative of the function, using the product rule:

`f'(x) = (x^(1/3))'(x+4) + (x^(1/3))(x+4)'`

`f'(x) = (x+4)/(3x^(2/3)) + (x^(1/3))`

You need to solve for x the equation f'(x) =0:

`(x+4)/(3x^(2/3)) + (x^(1/3)) = 0`

`x + 4 + (3x^(2/3))(x^(1/3)) = 0`

`x + 4 + 3x = 0 => 4x + 4 = 0 => x = -1`

**Hence, f'(x)<0 and the function decreases for `x in (-oo,-1) ` and f'(x)>0, the function increases for `x in (-1,oo)` .**

b) **Since for x = -1, f'(-1) = 0 and considering the monotony of the function, yields that the function has minimum point at x = -1.**

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