This is a question found in an A-Level math book.
Given a curve
$$f(x) = \frac{\cos2x}{e^x}, 0\le x\le\pi$$
Determine the interval where $f(x)$ is concave.
According to the math book:
The function $f(x)$ is concave on a given interval if and only if $f''(x) \le 0$ for every $x$ in that interval.
Also
The point at whitch a curve changes from being concave to convex is called a point of inflection.
A point of inflection is a point at which $f''(x)$ changes sign.
Hence, to calculate the interval, I first calculates the second derivative of $f(x)$:
$$f''(x)=\frac{4\sin2x-3\cos2x}{e^x}$$
From this point and onwards, I can either use the definition of concavity or the definition of the point of inflection to find the interval.
Using the definition of concavity
Set $f''(x) \le 0$:
$$\therefore\frac{4\sin2x-3\cos2x}{e^x} \le 0 $$
$$\because e^x > 0$$
$$\therefore4\sin2x-3\cos2x \le 0$$
$$\therefore\tan2x \le 0.75$$
Hence, the answer is
\begin{align}
0\le &x \le-0.322\\
\pi/4\le &x \le1.892\\
0.75\pi\le &x\le\pi\\
\end{align}
Using the definiton of point of inflection
The point of inflection is at $x=0.322$ and $x=1.892$.
And from the first derivative, one can find a local minimal at $x=1.34$ meaning the curve between the two point of inflection is convex.
Hence, the answer is
\begin{align}
0\le &x \le-0.322\\
1.892\le &x\le\pi\\
\end{align}
Which one is correct? And why is the other incorrect?
Update
I think both approaches are valid, but a mistake is produced when I divide $sin$ by $cos$ in the first answer. Is it because $\cos2x < 0$ when $0.25\pi< x <0.75\pi$, so $\le$ needs to be changed to $\ge$? i.e.
\begin{cases}
\tan2x \le 0.75, &0\le x<0.25\pi \text{ and } 0.75< x<\pi\\
\tan2x \ge 0.75, &0.25\pi <x<0.75\pi\\
\end{cases}
Solving this gives the second answer.
Best Answer
The correct answer is \begin{align} 0\le &x \le0.322\\ 1.892\le &x\le\pi\\ \end{align}
A proper way to solve the inequity,
$$4\sin2x-3\cos2x \le 0$$
is to follow the steps below,
$$5 \left( \sin2x\cdot \frac 45 - \cos2x\cdot \frac 35 \right) \le 0$$
$$ \sin2x \cos\theta - \cos2x \sin\theta \le 0$$
where $\theta = \arccos(4/5) = 0.6435$. Thus,
$$\sin(2x-0.6435)\le0$$
Then,
$$-\pi\le 2x-0.6425\le0, \>\>\> \pi\le 2x-0.6425 \le 2\pi$$
Thus, with the given range $0\le x\le \pi$, the solution is,
$$ 0\le x \le0.322, \>\>\> 1.892\le x\le\pi $$
which can also be verified from the attached plot.