Question

Find the points of inflection of f(x)=2sinx + sin2x.

I have the 2nd derivative: f''(x) = -2sinx - 4sin(2x)

Answer 1

ok... so you need to solve the 2nd derivative....

so let f"(x) = 0

and you might look at the fact that sin(2x) = 2sin(x)cos(x) as a substitution

Answer 2

Are my steps correct?

-2sin(x) - 4sin(2x) = 0
-2sin(x) - 8sinxcosx =0
-2sin(x)(1+4cosx)=0

-2sin(x) = 0
Point of inflection at pi, on the interval of [0,2pi]

4cosx=-1
^not so sure about this one.

Answer 3

well start with the point of inflexion at pi... what type of point of inflexion is it....

you needed to find the stationary points... solutions to the 1st derivative..

the other solution


cos(x) = -1/4

x = arccos(-1/4)

express the answer in radians

Answer 4

and don't forget over the interval [0, 2pi]

there will be points of inflexion at 0 and 2pi