Problem source

A function $\f(x)$ is said to be \textit{convex} in the interval $a < x < b$ if $\f''(x)\ge0$ for all $x$ in this interval.
\begin{questionparts}
\item Sketch on the same axes the graphs of $y= \frac23 \cos^2 x$ and $y=\sin x$ in the interval $0\le x \le 2\pi$.
The function $\f(x)$ is defined for $0 < x < 2\pi$ by
\[\f(x) = \e^{\frac23 \sin x}.
\]
Determine the intervals in which $\f(x)$ is convex.
 \item 
The function $\g(x)$ is defined for $0 < x < \frac12\pi$ by
\[\g(x) = \e^{-k \tan x}.
\]
If $k=\sin 2 \alpha$ and $0 < \alpha < \frac{1}{4}\pi$, show that $\g(x)$ is convex in the interval $0 < x < \alpha$, and give one other interval in which $\g(x)$ is convex.
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2/3*cos(deg(#1))^2};
    \def\functiong(#1){sin(deg(#1))};
    \def\xl{-1}; 
    \def\xu{7};
    \def\yl{-1.5}; \def\yu{2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid[xstep={pi/3}, ystep=.25] (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);
        \draw[curveA, domain=0:{2*pi}, samples=150] plot(\x, {\functionf(\x)});
        \draw[curveB, domain=0:{2*pi}, samples=150] plot(\x, {\functiong(\x)});

    \end{scope}
    
        \node[curveB, above] at ({pi/2}, 1) {$y = \sin x$};
        \node[curveA, above] at ({2*pi}, {2/3}) {$y = \frac23\cos^2 x$};
        
    \end{tikzpicture}
\end{center}


\begin{align*}
&& f(x) &= \exp\left (\tfrac23\sin x \right) \\
&& f'(x) &= \exp\left (\tfrac23\sin x \right) \cdot \tfrac23 \cos x  \\
&& f''(x) &= \left ( \exp\left (\tfrac23\sin x \right) \cdot \tfrac23\right) \left ( \tfrac23 \cos^2 x - \sin x \right)
\end{align*}

Therefore $f(x)$ is convex when $\frac23 \cos^2 x \geq \sin x$.

Note that we can find the equality points when
\begin{align*}
&& \sin x &= \frac23 \cos^2 x \\
&&&= \frac23 (1- \sin^2 x) \\
\Rightarrow && 0 &= 2\sin^2 x + 3 \sin x - 2 \\
&&&= (2 \sin x -1) (\sin x+2)
\end{align*}

ie $\sin x = \frac12 \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$.

Therefore $f$ is convex on $[0, \frac{\pi}{6}] \cup [\frac{5\pi}{6}, 2\pi]$

\item Suppose $g(x) = \exp \left ( -k \tan x \right)$ then
\begin{align*}
&& g'(x) &= \exp \left ( -k \tan x \right) \cdot (-k \sec^2 x ) \\
&& g''(x) &= \left ( -k \exp \left ( -k \tan x \right) \right) \left ( -k\sec^4 x + 2 \sec x \cdot \sec x \tan x\right) \\
&&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k  + 2\sin x \cos x \right) \\
&&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k  + \sin 2x  \right) \\
\end{align*}

If $0 < \alpha < \frac{\pi}{4}$ then $k > 0$ so $g$ is convex if $-k + \sin 2x < 0$, ie $\sin 2x < \sin 2\alpha$, ie on $(0, \alpha)$ and $(\frac{\pi}{2} - \alpha, \frac{\pi}{2})$

\end{questionparts}