How to find when a function is rising and decreasing I have this assignment that i don't fully understand.
$$f(x) = e^x\sin(x)$$ $$ x \in [0,2\pi]$$
1) decide where the function rises and decreases
2) find all the local maxima and minima and the global maxima and minima
3) decide where the function "curls" up and down
I do not really understand the task, so if someone could explain it will be appreciated.
 A: HINT
As you rightly observe in the comments, it is somewhat poorly worded but should make sense.
Concerning $iii)$ (I think $i), ii)$ are pretty clear), it asks you to find where the function is concave and where it is convex. 
We go about it using the second derivative.
$$f'(x)=e^x\sin x+e^x\cos=e^x(\sin x+\cos x) \Rightarrow\\f''(x)=e^x(\sin x+\cos x)+e^x(\cos x-\sin x)=2e^x\cos x$$
So you need to find the roots of $f''(x)$ in $[0,2\pi]$, that is $\xi\in [0,2\pi]$ such that $f''(\xi)=2e^{\xi}\cos \xi=0$ or equivalently $\cos \xi=0$.
Once you do that you have to make the table of signs to deduce concavity.
A: Studying the intervals where the function increases and decreases is done with the derivative, because a function that is strictly increasing over an interval has nonnegative derivative and, conversely, a function having positive derivative over an interval is strictly increasing.
Note that “increasing” is more common than “rising”.
In your case the derivative is, by the product rule,
$$
f'(x)=e^x\sin x+e^x\cos x=e^x(\sin x+\cos x)=
\sqrt{2}e^x\sin\left(x+\frac{\pi}{4}\right)
$$
Thus we have $f'(x)>0$ for $x\in(0,5\pi/4)$ or $x\in(7\pi/4,2\pi)$.
Therefore we can conclude that


*

*$f$ is increasing over $[0,5\pi/4]$

*$f$ is decreasing over $[5\pi/4,7\pi/4]$

*$f$ is increasing over $[7\pi/4,2\pi]$
Consequently, $0$ is a local minimum, $5\pi/4$ is a local maximum, $7\pi/4$ is a local minimum and $2\pi$ is a local maximum.
The absolute maximum and minimum are among the local ones. Since


*

*$f(0)=0$

*$f(5\pi/4)=\frac{e^{5\pi/4}}{\sqrt{2}}$

*$f(7\pi/4)=-\frac{e^{5\pi/4}}{\sqrt{2}}$

*$f(2\pi)=0$
you can immediately spot the absolute maximum and minimum.
For the “curling up” and “curling down”, which are usually called “convexity” and “concavity”, you need to compute the second derivative; positive sign implies convexity:
$$
f''(x)=e^x\sin x+e^x\cos x+e^x\cos x-e^x\sin x=2e^x\cos x
$$
which is positive over $(0,\pi/2)$ or over $(3\pi/2,2\pi)$.
The inflection points are this at $\pi/2$ and $3\pi/2$, where there is a change between convexity and concavity.
In the graph below I divided the function by $20$ in order not to have big values.

