A. find the critical points of the following function on the given interval

b. use a graphing utility to determine whether the critical points correspond to local maxima, and local minima, or neither.

c. find the absolute maximum and minimum values on the given interval.

f(theta)=sin(theta)+4 cos(theta); [-2pi,2pi]

identify all the critical points.

identify all the critical points that are local minima.

A) Given function:

to determine the critical points we need to differentiate and equate it to zero. (the slopes of the curve at the critical point is always zero)

now use

now solve for within the range

as positive tan lies in the first and third quadrant of the unit-circle. our values within the interval will be:

These are the critical points!

We can use these values to find the values of

B) From the graph we can see that the first and third are maxima and second and fourth are minima.

C) Since all minimum points have the same y-coordinate, and all maximum points have the same y-coordinate. We can safely say that all points are local critical points in this function.

yes it is.

diagonals of parallelogram bisect each other so if the diagonals are of equal length then they bisect each other and length of each bisect is equal. so the angle made by these bisects with any sides should be equal ( it is a theorem of geometry ) if all such angles are equal then the length all the sides with which these angles are made should of equal.now if all the sides has same length then it is rhombus and here all angles have same values that is x+x+x+x=360; x=90

so it is proved that it is a rectangle.