Given the function f(x) = -x^4 + 12x^3 - 36x^2, determine all intervals on which f is increasing.

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To determine where the function f(x) = -x^4 + 12x^3 - 36x^2 is increasing, we need to find the derivative f'(x) and determine where it is positive. Find the derivative: f'(x) = d/dx (-x^4 + 12x^3 - 36x^2) = -4x^3 + 36x^2 - 72x. Factor the derivative: f'(x) = -4x(x^2 - 9x + 18). Factor further: f'(x) = -4x(x - 3)(x - 6). Determine the critical points by setting f'(x) = 0: -4x(x - 3)(x - 6) = 0, which gives x = 0, x = 3, and x = 6. Test intervals around the critical points to determine where f'(x) is positive: For x in (-∞, 0), choose x = -1: f'(-1) = -4(-1)((-1) - 3)((-1) - 6) = -4(-1)(-4)(-7) = -112 (negative). For x in (0, 3), choose x = 1: f'(1) = -4(1)(1 - 3)(1 - 6) = -4(1)(-2)(-5) = -40 (negative). For x in (3, 6), choose x = 4: f'(4) = -4(4)(4 - 3)(4 - 6) = -4(4)(1)(-2) = 32 (positive). For x in (6, ∞), choose x = 7: f'(7) = -4(7)(7 - 3)(7 - 6) = -4(7)(4)(1) = -112 (negative). Therefore, the function is increasing on the interval (3, 6).

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Given the function f(x) = -x^4 + 12x^3 - 36x^2, determine all intervals on which f is increasing.

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