Find the least and greatest values of f(x) = e^xsinxf(x)=exsinx for 0<=x<=pi0xπ ?

1 Answer
Jun 18, 2018

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}
f(x): x in (0,pi)f(x):x(0,π) has a maximum value of approx 7.467.46 at x = (3pi)/4x=3π4

Explanation:

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