The second derivative at a maximum can be zero, as demonstrated by the function f(x) = -x^4, where f'(0) = f''(0) = f'''(0) = 0 and f^{(iv)}(0) < 0. This indicates that while the gradient transitions from positive to negative around the maximum, the second derivative does not necessarily have to be negative. The discussion clarifies that a plateau at the maximum does not invalidate it as a point of interest in calculus.
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Mathsforum100 said:The second derivative at a maximum is either negative or zero. Can you explain how it can be zero? There can't be a 'plateau' at the maximum or it would not be a point. I cannot imagine graphically how the second derivative at a maximum can be zero. Before the maximum, the gradient is positive. After the maximum it is negative. So the gradient is decreasing.