For the function, use the second derivative test (if possible) to determine if each critical point is a minimum, maximum, or neither. If the second derivative test can't be used, say so.
f (x) = exsin x for 0 < x < 2π
The first derivative is
f'(x) = excos x + exsin x = ex(cos x + sin x).
This is never undefined. Since ex is never 0, f ' is only 0 when
cos x + sin x = 0,
cos x = -sin x.
On the interval 0 < x < 2π, this only occurs at and , so these are our critical points. The second derivative is
f "(x) = ex(-sin x) + excos x + excos x + exsin x = 2excos x.
Now we need to find the sign of the second derivative at each critical point. Since 2ex is always positive, the sign of f "(x) is the same as the sign of cos x.
Since is negative, we know
so f is concave down and has a maximum at . Since is positive, we know so f is concave up and has a minimum at .
A graph of the function supports these answers: