Question

Question #cf6e2

Answer 1

See below.

Explanation:

Calling `f(x) = (1/2+x)(1/3-x) = -x^2-x/6+1/6` and

we know that the local minima/maxima are given at the stationary points or when

`d/(dx)f(x) = 0`

but in our case we have

`f(g(x))` so the stationary points are when

`d/(dx)f(g(x))=-1/6(1+12 g(x))g'(x)=0`

with `g(x) = cos(x)`

so the stationary points are the solutions of

`{(1+12g(x)=1+12cos(x)=0),(g'(x)=-sin(x)=0):}`

and the stationary points are

`x_1 = -"arccos"(-1/12)+2kpi` and

`x_2 = k pi` for `k in ZZ`

The stationary points qualification is done using `d^2/(dx^2)f(g(x))`

or

`d^2/(dx^2)f(g(x))=-2g'(x)^2 - 1/6 (1 + 12 g(x)) g''(x)`

or

`d^2/(dx^2)f(cos(x))=Cos(x)/6 + 2 Cos^2(x) - 2 Sin^2(x)`

Regarding the stationary points the value of `d^2/(dx^2)f(cos(x))`qualifies the type of extremum. If `d^2/(dx^2)f(cos(x)) > 0` is a local minimum and if `d^2/(dx^2)f(cos(x)) < 0` is a local maximum.

so

`d^2/(dx^2)f(cos(x_1)) = -143/72` local maxima
`d^2/(dx^2)f(cos(x_2)) = 13/6` local minima

Attached a plot showing this behavior.

so concluding, `x_1` for the maxima and `x_2` for the minima

NOTE. The same result for the maximum point would be obtained directly by making

`(df)/(dx)=-1/6 - 2 x=0->x=-1/12` with the maximum value given by

`f(-1/12)=25/144` and here

`(d^2f)/(dx^2)(-1/12)=-2` qualifying a maximum.

Answer 2

Maximum value is `25/144`

Explanation:

Let:

`y = (cosx+1/2)(1/3-cosx)`
`\ \ = 1/3cosx-cos^2x+1/6 -1/2cosx`
`\ \ = 1/6-cos^2x+ -1/6cosx`

This is a quadratic in `cosx` so let us complete the square

`y = -(cos^2x+1/6cosx-1/6)`
`\ \ = -((cosx+1/12)^2-(1/12)^2-1/6)`
`\ \ = -((cosx+1/12)^2-25/144)`
`\ \ = 25/144 - (cosx+1/12)^2`

Now `(cosx+1/12)^2 ge 0 AA x in RR`, and so `y` has a maximum when `(cosx+1/12)^2 = 0 => cosx=-1/12`.

When this occurs, the maximum value of `y` is achieved.

`y_(max) = 25/144 - 0 = 25/144`