Question

How do you find all local maximum and minimum points using the second derivative test given `y=sin^3x`?

Answer 1

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, ...`

Explanation:

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