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

1 Answer
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]}