For the function `f(x)=x^(1/3)(x+20)` find where f(x) is increasing and decreasing

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You need to evaluate the derivative of the function to tell where the function increases or decreases, such that:

`f'(x) = (x^(1/3))'(x + 20) + (x^(1/3))(x + 20)'`

`f'(x) = (1/3)*x^(1/3 - 1)(x + 20) + x^(1/3)`

`f'(x) = (1/3)*x^(-2/3)(x + 20) + x^(1/3)`

You need to set the equation `f'(x) = (1/3)*x^(-2/3)(x + 20) + x^(1/3)` equal to 0, such that:

`(1/3)*x^(-2/3)(x + 20) + x^(1/3) = 0 => (1/3)*x^(-2/3)(x + 20) = -x^(1/3)`

`(x + 20)/(3x^(2/3)) = -x^(1/3) => x + 20 = -3x^(2/3+1/3)`

`x + 20 = -3x => 3x + x = -20 => 4x = -20 => x = -5`

**You need to notice that the function decreases over the interval `(-oo,-5)` and it increases over the interval **`(-5,oo).`

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