Discussion
At x = 0, the function \(f(x)=x^{3}+1\) has
Answer: D
At x = 0, \(x^{3}+1\) has a point of inflection.
The first derivative \(f'(x) = 3x^{2}\) is zero at x = 0, so it is a stationary point. But \(f'(x)\) is positive on both sides, so the function keeps increasing through x = 0 and there is no maximum or minimum. The second derivative \(f''(x)=6x\) changes sign from negative to positive there, so the curve changes from concave down to concave up: a stationary point of inflection.
A singularity would need the function or its derivative to blow up, but a polynomial is smooth everywhere.
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