# How do you sketch the graph y=(2+sinx)^2 using the first and second derivatives from 0<=x<2pi?

##### 1 Answer
Nov 19, 2016

See graph and explanation. x is in radian measure. $\pi$ measures 3.1 units, nearly. The period of y is $2 \pi = 6.25$ units, nearly.

#### Explanation:

graph{ y= (2+sin x)^2 [-20, 20, -10, 10]} The period of y is $2 \pi$.

As 2 + sin x ranges from 1 to 3, y oscillates between 1 and 9.

y' = 2(2+sin x) cos x =0, when cos x = 0 and this x = an odd multiple

of $\frac{\pi}{2} = 1.57$, giving turning points in the waves.

y''

$= - 2 \sin x \left(2 + \sin x\right) + 2 {\cos}^{2} x$

$= - 4 {\sin}^{2} x - 4 \sin x + 2 = 0$, when

$\sin x = \frac{\sqrt{3} - 1}{2}$ giving $x = k \pi + {\left(- 1\right)}^{k} \left(0.375\right)$ radian, k = 0, +-1,

+-2, +-3, ..., giving points of inflexion (POI), upon the axial line y = 5.

I would like the readers to read my answer once again, looking at

the graph after reading every line in my answer.

The graph asked for is between x = 0 and x = 6.28, nearly, with ends

at (0, 4) and (6.28, 4).