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*) = *e*^{x}sin *x* for 0 < *x* < 2π

Answer

The first derivative is

*f*'(*x*) = *e*^{x}cos *x* + *e*^{x}sin *x* = *e*^{x}(cos *x* + sin *x*).

This is never undefined. Since *e*^{x} is never 0, *f *' is only 0 when

cos *x* + sin *x* = 0,

equivalently,

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*) = *e*^{x}(-sin *x*) + *e*^{x}cos *x* + *e*^{x}cos *x* + *e*^{x}sin *x* = 2*e*^{x}cos *x*.

Now we need to find the sign of the second derivative at each critical point. Since 2*e*^{x} 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: