`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.
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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).
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