Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 9

A stationary point is a point where the function is differentiable and its first derivative vanishes. Interpret the cube root as the real cube root, and put $q(x)=x^2-8x+7$. Then $f(x)=15+(x-3)\sqrt[3]{q(x)}$. For $x\ne 1,7$, the product rule and chain rule giveThe denominator is positive wherever it is defined. Consequently the two stationary abscissae areUsing the stationary equation to simplify the value of $q$ gives $q(x_{\pm })=-2(x_{\pm }+5)/5$. Thus the complete stationary points areNumerically they are approximately $(1.46762,17.10362)$ and $(6.13238,9.84664)$. The numerator of $f^{\prime }$ is positive outside $[x_{-},x_{+}]$ and negative inside, so the first is a local maximum and the second a local minimum.

At $x=1,7$, the real function is continuous but has a vertical tangent; the factor $x-3$ does not cancel either cube-root singularity. Those points are not stationary points. This is the distinction required when finding stationary points with real cube-root factors.

Differentiate to obtain $g^{\prime }(y)=-1-4\sin y$. Let $\alpha =\arcsin (1/4)$, with $0<\alpha <\pi /2$. All solutions of $g^{\prime }=0$ areSince $g^{\prime \prime }(y)=-4\cos y$, the first family consists of local maxima and the second of local minima. Their function values follow from $4\cos \alpha =\sqrt{15}$:The derivative is periodic, but $g$ itself is not: increasing $y$ by $2\pi$ lowers $g$ by $2\pi$, which accounts for the indexed vertical offsets.

For $|x|<1$, the geometric series gives $(1+x{)}^{-1}={\sum }_{j=0}^{\infty }(-x{)}^j$. Integrating from zero to $x$ yields the Maclaurin seriesOn the real axis the series also converges at $x=1$ to $\ln 2$, but diverges at $x=-1$.

For fixed real $a$ and sufficiently large positive $x$, the base in the requested limit is positive. Apply the series with argument $-a^2/x^2$:Continuity of the exponential function therefore givesThis includes $a=0$, when the expression is identically one.

Set $h=x+3$, so $x=-3(1-h/3)$. The negative binomial series gives $(1-u{)}^{-4}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+3}{3})u^n$ for $|u|<1$. Hence the exact Taylor series isThe initial terms are $\frac{7}{81}[1+4h/3+10h^2/9+20h^3/27+\cdots ]$. Equivalently the coefficient of $(x+3{)}^n$ is $7(n+3)!/[6n!3^{n+4}]$, obtained by differentiating $7x^{-4}$ repeatedly and evaluating at $-3$.

The radius of convergence is three, set by the pole at $x=0$. At either endpoint $h=\pm 3$, the terms fail to tend to zero, so neither endpoint is included.