#f(x) = e^xsinx#
Firstly, #f(x):x in [0,pi]# has a trivial minimum value of #0# for #x={0,pi}#
Now lets consider #f(x) = e^xsinx: x in (0,2pi)#
To find extrema: #f'(x) =0#
#f'(x) = e^xcosx+e^xsinx# [Product rule]
So, to find extrema: #e^xcosx+e^xsinx =0#
#e^x(cosx+sinx)=0#
Since #e^x !=0 -> cosx+sinx =0#
Now, #cosx=-sinx# for #x ={(3pi)/4, (7pi)/4} in (0,2pi)#
Hence, extrema of #f(x) in (0,2pi)# are:
# f((3pi)/4) = e^((3pi)/4)sin((3pi)/4)#
#approx 10.551 xx 0.707 approx 7.46#
#f((7pi)/4) = e^((7pi)/4)sin((7pi)/4)#
#approx 244.151 xx -0.707 approx -172.64#
Thus:
#f(x)_max approx 7.46: x = (3pi]/4#
#f(x)_min approx -172.64: x = (7pi]/4#
Therefore #f(x)# for #x in (0,pi) # has a maximum value of #approx 7.46# at #x = (3pi)/4#
