Problem source

The function $\mathrm{f}$ and $\mathrm{g}$ are related (for all real $x$) by 
\[
\mathrm{g}(x)=\mathrm{f}(x)+\frac{1}{\mathrm{f}(x)}\,.
\]
Express $\mathrm{g}'(x)$ and $\mathrm{g}''(x)$ in terms of $\mathrm{f}(x)$ and its derivatives. 
If $\mathrm{f}(x)=4+\cos2x+2\sin x$, find the stationary points of $\mathrm{g}$ for $0\leqslant x\leqslant2\pi,$ and determine which are maxima and which are minima.

Solution source

$g'(x) = f'(x) - \frac{f'(x)}{(f(x))^2} = f'(x) \l 1 - \frac{1}{(f(x))^2} \r$

$g''(x) = f''(x) - \frac{f''(x)f(x)^2-f'(x)\cdot 2f(x) f'(x)}{(f(x))^4} = f''(x) + \frac{f''(x)f(x)-2(f'(x))^2}{(f(x))^3}$

\begin{align*}
f(x) &=4+\cos2x+2\sin x \\
f'(x) &=-2\sin2x+2\cos x \\
f''(x) &= -4\cos2x-2\sin x
\end{align*}

Therefore, since the stationary points of $g$, ie points where $g'(x) = 0$ are where $f'(x) = 0$ or $f(x) = \pm 1$ we should look at

\begin{align*}
&& 0 &= f'(x) \\
&& 0 &= 2 \cos x - 2 \sin 2x \\
&&&= 2 \cos x - 4 \sin x \cos x \\
&&&= 2\cos x (1 - 2 \sin x) \\
\Rightarrow && x &= \frac{\pi}2, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}
\end{align*}

\begin{align*}
&& 1 &= f(x) \\
&& 1 &= 4 + \cos 2x + 2 \sin x \\
\Rightarrow && \cos 2x = -1,& \sin x = -1 \\
\Rightarrow && x &= \frac{3\pi}{2}
\end{align*}

which we were already checking.

For each of these points we have:

\begin{array}{c|c|c|c||c}
x & f(x) & f'(x) & f''(x) & g''(x) \\ \hline
\frac{\pi}{2} & 5 & 0 & 2 & > 0\\
\frac{3\pi}{2} & 1 & 0 & 6 &> 0\\
\frac{\pi}{6} & 5.5 & 0 & -3 & < 0 \\
\frac{5\pi}{6} & 5.5 & 0 & -3 & < 0\\
\end{array}

Therefore $\frac{\pi}{2}, \frac{3\pi}{2}$ are minimums and $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ are maxima.