Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 6 / b

The derivative factors asThe polynomial is differentiable everywhere, so its critical points occur precisely where this derivative vanishes. HenceTheir values are $g(0)=5$ and $g(3)=32$.

The relevant derivatives at $x=3$ areThus the cubic Taylor polynomial isIn fact the exact expansion is $g(x)=32+4(x-3{)}^3+(x-3{)}^4$, so the omitted term has order $O((x-3{)}^4)$ in big O notation.

At $x=0$, $g^{\prime \prime }(0)=36>0$, so the second-derivative test gives a strict local minimum. In fact $g^{\prime }(x)=4x(x-3{)}^2$ is negative for $x<0$ and positive for $x>0$ except at $x=3$, so $g(0)=5$ is the global minimum as well.

At $x=3$, the second-derivative test is inconclusive because $g^{\prime \prime }(3)=0$. But $g^{\prime \prime }(x)=12(x-1)(x-3)$ changes from negative to positive across $3$, while $g^{\prime }(3)=0$. Hence this is a stationary inflection point. Equivalently, the leading nonconstant term $4(x-3{)}^3$ in the Taylor polynomial changes sign across $3$. The paper explicitly uses “saddle point” to mean an inflection point, so in its terminology the classification isHere the “saddle point” is a one-dimensional stationary inflection point, rather than the usual multivariable use of saddle point of a scalar function.^[ref]