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

Question

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

Emma Aldridge · Accepted Answer

Please see below.

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]}

Isabella Ackerman · Answer

undefined To sketch the graph $ y = \sin(x) + \sin^2(x) $ using the first and second derivatives:

1. Find the first derivative of $ y $ with respect to $ x $ and set it equal to zero to locate critical points.
2. Find the second derivative of $ y $ with respect to $ x $ to determine concavity.
3. Determine the behavior of $ y $ as $ x $ approaches critical points and at endpoints of the interval $ 0 \leq x < 2\pi $.
4. Use the information gathered to sketch the graph within the given interval.

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

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