We have # y =sin^3x # .... [1]

We first find the max/min critical points by finding values of #x# such that #dy/dx=0#

Differentiating [1] wrt #x# using the chain rule we get:

# dy/dx = 3sin^2xcosx # .... [2]

# dy/dx = 0 => 3sin^2xcosx = 0#

# :. sin^2xcosx = 0#

# :. sin^2x = 0 # or #cosx = 0#

If # sin^2x = 0 => sin x =0 => x=npi, n in ZZ #

And #cosx = 0 => x=npi-pi/2, n in ZZ #

To determine the nature of these points we need to look at the second derivative

Differentiating [2] wrt [x] and simultaneously applying the chain rule and product rule we have;

# (d^2y)/dx^2 = (3sin^2x)(-sinx) + (6sinxcosx)(cosx) #

# :. (d^2y)/dx^2 = 6sinxcos^2x - 3sin^3x #

We already know that min/max occurs when #x=npi# or# x=npi-pi/2#, so let's find # (d^2y)/dx^2# at these points;

When # x=npi => (d^2y)/dx^2 = 6(0)cos^2x - 3(0) = 0 #

So we can conclude that # x=npi # correspond to points of inflexion

When # x=npi-pi/2 => (d^2y)/dx^2 = 6sinx(0) - 3sin^3(npi-pi/2) #

# :. (d^2y)/dx^2 = - 3sin^3(npi-pi/2) # .... [3]

Now # sin(A-B) -= sinAcosB - cosAsinB #

# :. sin(npi-pi/2) = sinnpicos(pi/2) - cos npi sin (pi/2) #

# :. sin(npi-pi/2) = (sinnpi)(0) - cos npi (1) #

# :. sin(npi-pi/2) = - cos npi #

And so, substituting into [3] gives us:

# :. (d^2y)/dx^2 = - 3(- cos npi)^3 #

# :. (d^2y)/dx^2 = 3cos^3 npi #

So we can conclude that

# { ((d^2y)/dx^2 <0, n " odd",=>, "maximum"), ((d^2y)/dx^2>0, n " even", =>, "minimum") :} #

So Points of Inflection occur when #x=npi#,

ie #x=-3pi,-2pi,-pi,0,pi,2pi,3pi, ... #

Maximum points occur when #x=npi-pi/2# and n is odd

ie #x=(-11pi)/2,(-7pi)/2,(-3pi)/2,pi/2, (5pi)/2, (9pi)/2, ...#

Minimum points occur when #x=npi-pi/2# and n is even

#x=(-9pi)/2,(-5pi)/2,(-pi/2), (3pi)/2,(7pi)/2, ...#

This can be visualised by the graph