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Near zero, $x^3-x^5=x^3(1-x^2)$ has the sign of $x$, so $f(x)<1$ just to the left and $f(x)>1$ just to the right. Thus the critical point is neither a local minimum nor a local maximum. Moreover the Taylor series gives $f^{\prime \prime }(x)=6x+O(x^3)$, which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.

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]