Problem source

Find the stationary points of the function $\mathrm{f}$ given by
\[
\mathrm{f}(x)=\mathrm{e}^{ax}\cos bx,\mbox{ }(a>0,b>0).
\]
Show that the values of $\mathrm{f}$ at the stationary points with $x>0$ form a geometric progression with common ratio $-\mathrm{e}^{a\pi/b}$. 
Give a rough sketch of the graph of $\mathrm{f}$. 

Solution source

Let $f(x) = e^{ax} \cos bx$ then, $f'(x) = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax} \l a\cos bx - b \sin bx \r$. Therefore the stationary points are where $f'(x) = 0 \Leftrightarrow \tan bx = \frac{b}a$, ie $x = \tan^{-1} \frac{a}{b} + \frac{n}{b} \pi, n \in \mathbb{Z}$.

\begin{align*}
f(\tan^{-1} \frac{a}{b} + \frac{n}{b} \pi) &= e^{a \tan^{-1} \frac{a}{b} + \frac{an}{b} \pi} \cos \l b \tan^{-1} \frac{a}{b} +n \pi\r \\
&= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot e^{\frac{an}{b} \pi}(-1)^n \\
&= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot (-e^{\frac{a}{b} \pi})^n \\
\end{align*}

showing the form the desired geometric progression.

\begin{center}
\begin{tikzpicture}[scale=1]
    \def\e{2.7182}
    \def\a{0.5}
    \def\b{1.9}
    \def\p{3.141592}
    \def\s{8}
    \draw[->] (-1,0) -- (8,0);
    \draw[->] (0,-3) -- (0,3);
    \draw[domain = -1:\s, samples=360, variable = \x]  plot ({\x},{\e^(\a*\x - \s*\a + 1.1)*cos(\b*\x*180/\p)});
    \draw[domain = -1:\s, samples=180, variable = \x, dashed]  plot ({\x},{-\e^(\a*\x - \s*\a + 1.1)});
    \draw[domain = -1:\s, samples=180, variable = \x, dashed]  plot ({\x},{\e^(\a*\x - \s*\a + 1.1)});

    \node at (8,0) [right] {$x$};
    \node at (0,3) [above] {$y$};

    \node at (8,3.1) [right] {$y = e^{ax}$};
    \node at (8,-3.1) [right] {$y = e^{-ax}$};
    \node at (8,-2) [right] {$y = e^{-ax} \cos bx$};
\end{tikzpicture}
\end{center}