Question #cf6e2

2 Answers
Apr 7, 2017

Answer:

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.

enter image source here

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.

Jun 14, 2017

Answer:

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 #