Question

How do you find all local maximum and minimum points using the second derivative test given `y=cos^2x-sin^2x`?

Answer 1

`y(x)` has a local maximum for `x=kπ` and a local minimum for `x=kπ+π/2`

Explanation:

As:

`y=cos^2x-sin^2x=cos(2x)`

we do not really need derivatives because we know that `cos(2x)` has a local maximum for `x=kpi` where `cos(2x)=1` and a local minimum for `x=kpi+pi/2` where `cos(2x)=-1`.

However we can check:

`y'(x) = -2sin(2x)`

so the critical points occur for:

`2x= kpi` or `x=kpi/2`

`y''(x)= -4cos(2x)`

and in fact:

`y''(2kpi) = -4 < 0` so these points are maximums, and

`y''((2k+1pi)) = 4 > 0` so these points are minimums.