Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 5 / a

The chain rule givesThe exponential function never vanishes, so all critical points areIn particular $f^{\prime }(0)=0$.

Use the Taylor series $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the Taylor series isThe linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in big O notation, follows because the fourth-order term is also zero; the next term is $-x^5$.

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.