Problems from Maximun, minimum and inflection points of a function

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: $$y^{\prime }=cos(x)=0\Rightarrow x=\pm {\displaystyle \frac{\pi }{2}},\pm {\displaystyle \frac{3\pi }{2}},\pm {\displaystyle \frac{5\pi }{2}},\pm {\displaystyle \frac{7\pi }{2}},\dots$$

We compute the sign of each of the solutions to determine if it is a maximum or a minimum: $y^{″}=-sin(x)$. $$x={\displaystyle \frac{\pi }{2}}\to y^{″}({\displaystyle \frac{\pi }{2}})=-1<0\Rightarrow Max$$ $$x={\displaystyle \frac{3\pi }{2}}\to y^{″}({\displaystyle \frac{3\pi }{2}})=1>0\Rightarrow Min$$ $$x={\displaystyle \frac{5\pi }{2}}\to y^{″}({\displaystyle \frac{5\pi }{2}})=-1<0\Rightarrow Max$$ $$\dots$$

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: $$y^{″}=-sin(x)=0\Rightarrow x=0,\pm \pi ,\pm 2\pi ,\pm 3\pi ,\dots$$ $$y(x)=sin(x)=0$$

See now the graph:

Solution:

Maxima: $({\displaystyle \frac{\pi }{2}},1),({\displaystyle \frac{5\pi }{2}},1),({\displaystyle \frac{9\pi }{2}},1),\dots$

Minima: $({\displaystyle \frac{3\pi }{2}},-1),({\displaystyle \frac{7\pi }{2}},-1),({\displaystyle \frac{11\pi }{2}},-1),\dots$

Inflection points: $(0,0),(\pm \pi ,0),(\pm 2\pi ,0),\dots$

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