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

1 Answer
Shwetank Mauria
Jun 12, 2017

Please see below.

Explanation:

We have y=sinx+sin^2x=sinx(1+sinx)

Observe that y=0 when sinx=0 or 1+sinx=0

hence between 0 <= x <= 2pi, it intersects x-axis at x={0,pi,(3pi)/2,2pi} or {0,3.1416,4.7124,6.2832}

Now we have extrema when (dy)/(dx) and as

(dy)/(dx)=cosx+2sinxcosx=cosx(1+2sinx) or cosx+sin2x

also (d^2y)/(dx^2)=-sinx+2cos2x

Hence extrema occurs when cosx=0 i.e. {pi/2,(3pi)/2}. While second derivative is negative at both these points, when 1+2sinx=0 i.e. sinx=-1/2 or x={(5pi)/3,(7pi)/3}, it is positive.

Hence we have a local minima at x={(5pi)/3,(7pi)/3} and local maxima at {pi/2,(3pi)/2}.

The graph appears as shown below.

graph{sinx(1+sinx) [-2.917, 7.083, -1.84, 3.16]}

Related questions
See all questions in Examples of Curve Sketching
Impact of this question
3499 views around the world
You can reuse this answer
Creative Commons License