Consider f(x) = (x^3)/3 - 4x^2 + 12x + 1. Find both the absolute max and absolute min of the function on the interval [3, 12).
Added by Ashley L.
Step 1
To find the critical points, we need to find the derivative of the function f(x) and set it equal to 0. f'(x) = x^2 - 8x + 12 Setting f'(x) = 0, we get: x^2 - 8x + 12 = 0 (x - 2)(x - 6) = 0 So, the critical points are x = 2 and x = 6. However, we need to check Show more…
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