Find the maxima, minima and inflection points of the function $$f(x)=sin(x)$$
Development:
We will solve the problem without having the graph of the sine.
Maxima / minimums.
First of all, we compute the derivative of the sine and we find its roots: $$$y'=cos(x)=0 \Rightarrow x=\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\pm\dfrac{5\pi}{2},\pm\dfrac{7\pi}{2},\ldots$$$
We compute the sign of each of the solutions to determine if it is a maximum or a minimum: $$y''=-sin(x)$$. $$$x=\dfrac{\pi}{2} \rightarrow y''(\dfrac{\pi}{2})=-1 < 0 \Rightarrow Max$$$ $$$x=\dfrac{3\pi}{2} \rightarrow y''(\dfrac{3\pi}{2})=1 > 0 \Rightarrow Min$$$ $$$x=\dfrac{5\pi}{2} \rightarrow y''(\dfrac{5\pi}{2})=-1 < 0 \Rightarrow Max$$$ $$$\ldots$$$
The values of the function in the maximum is $$1$$ and in the minimum is $$-1$$.
Inflection points.
The second derivative is equal to zero: $$$y''=-sin(x)=0 \Rightarrow x=0,\pm\pi,\pm2\pi,\pm3\pi,\ldots$$$ $$$y(x)=sin(x)=0$$$
See now the graph:
Solution:
Maxima: $$(\dfrac{\pi}{2},1),(\dfrac{5\pi}{2},1),(\dfrac{9\pi}{2},1),\ldots$$
Minima: $$(\dfrac{3\pi}{2},-1),(\dfrac{7\pi}{2},-1),(\dfrac{11\pi}{2},-1),\ldots$$
Inflection points: $$(0,0),(\pm\pi,0),(\pm2\pi,0),\ldots$$