The interval in which f(x) = x^3 – 3x + 2 increases is
Added by Victor C.
Step 1
To do this, we take the derivative of the function: f'(x) = 3x^2 – 3 Setting f'(x) = 0, we get: 3x^2 – 3 = 0 x^2 – 1 = 0 (x – 1)(x + 1) = 0 So the critical points are x = 1 and x = –1. Show more…
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