f(x) = e^xsinxf(x)=exsinx
Firstly, f(x):x in [0,pi]f(x):x∈[0,π] has a trivial minimum value of 00 for x={0,pi}x={0,π}
Now lets consider f(x) = e^xsinx: x in (0,2pi)f(x)=exsinx:x∈(0,2π)
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