Calculate the extrema for #f(x) = 3^(3 cos 2x+4 sin2x)# ?

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Answer 1 · 2017-06-18T20:35:14

See below.

We have #f(x) = 3^(3 cos 2x+4 sin2x)#

Considering #g(x)=3 cos 2x+4 sin2x# the stationary points are located at

#(dg)/(dx) = -12 Cos(2 x) - 16 Sin(2 x) = 0#

giving

#x = 1/2 (arctan(4/3)+2kpi), k=0,1,2,cdots# for the maxima points and
#x = 1/2(arctan(4/3)-pi+2kpi), k=0,1,2,cdots#for the minima points.

The qualification is made by computing

#(d^2g)/(dx^2)# at the stationary points. Positive values are associated with minima points and negative points to maxima points.

The evaluation is left to the reader.