Find the least and greatest values of #f(x) = e^xsinx# for #0<=x<=pi# ?

1 Answer
Jun 18, 2018

#f(x):x in [0,pi]# has a trivial minimum value of #0# for #x={0,pi}#
#f(x): x in (0,pi)# has a maximum value of #approx 7.46# at #x = (3pi)/4#

Explanation:

#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#