Let y=f(x)=(x^2-8 sqrt(-x))^2. find the intervals on which f is increasing/decreasing

Answer on assumption that f is dealt in real numbersf is increasing on the open interval (-4, -∛4) and f is decreasing on the open intervals (-∞, -4) and (-∛4, 0).======assuming the domain of x cannot contain complex numbers, we establish first that the domain is x  ≤ 0 due to the presence of  √(-x).to find where f is increasing/decreasing, we must find where the derivative of f is positive or negative. in order to do that, we must differentiate:     note that  find the critical numbers. these are where f' is undefined or equal to zero.we can see immediately that f' is undefined at x = 0 since there would be a division by zero.x = 0 is a critical number.to find where f' is equal to zero, set f'(x) = 0 and solve. in this case, we can use the zero factor property.    setting  to equal zero:     so x = -4 is a critical number.setting  to equal zero:     so x = -∛(4) is a critical number.our critical numbers are -4, -∛(4), and 0. setting up an interval chart to test for the sign of f' in the intervals created by these critical numbers, plug in a number that is in each interval to find the sign.    therefore, f is increasing on the open interval (-4, -∛4) and f is decreasing on the open intervals (-∞, -4) and (-∛4, 0).attached is a graph of function f from desmos to confirm

Idon’t understand what you’re asking..