# `f(x) = sin(x) cos(x) + 5` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema. This function is continuous and differentiable on the given interval. It is increasing when f'(x)>0 and decreasing when f'(x)<0. Let's find f'(x).

f(x) = sin(x)*cos(x) + 5 = (1/2)*sin(2x) + 5.

f'(x) = cos(2x).

It is zero at `2x=pi/2+k*pi` , or `x=pi/4+(k*pi)/2` . There are four such points on` (0,...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

This function is continuous and differentiable on the given interval. It is increasing when f'(x)>0 and decreasing when f'(x)<0. Let's find f'(x).

f(x) = sin(x)*cos(x) + 5 = (1/2)*sin(2x) + 5.

f'(x) = cos(2x).

It is zero at `2x=pi/2+k*pi` , or `x=pi/4+(k*pi)/2` . There are four such points on` (0, 2pi)` : `pi/4` , `(3pi)/4` , `(5pi)/4` and `(7pi)/4` .

f'(x) is positive on `(0, pi/4)` , (`(3pi)/4` , `(5pi)/4` ) and `((7pi)/4, 2pi)` , f is increasing on these intervals. f'(x) is negative on `(pi/4, (3pi)/4)` and `((5pi)/4, (7pi)/4)` , f is decreasing on these intervals.

Therefore the relative extrema are `pi/4` (maximum), `(3pi)/4` (minimum), `(5pi)/4` (maximum) and `(7pi)/4` (minimum).

Approved by eNotes Editorial Team