Question

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

Answer 1

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

Explanation:

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

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 `pi/2=1.57`, giving turning points in the waves.

y''

`=-2 sin x (2 + sin x)+2 cos^2x`

`=-4 sin^2x-4sinx+2=0`, when

`sin x=(sqrt3-1)/2` giving `x =kpi+(-1)^k(0.375)` 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).