What are the absolute extrema of # f(x)= |sin(x) - cos(x)|# on the interval [-pi,pi]? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer A. S. Adikesavan Apr 3, 2016 0 and #sqrt2#. Explanation: #0<=|sin theta|<=1# #sin x - cos x = sin x -sin (pi/2-x) =2 cos ((x+pi/2-x)/2) sin ((x-(pi/2-x))/2) =- 2 cos (pi/4) sin (x-pi/4) = -sqrt2 sin (x-pi/4)# so, #|sin x - cos x|=| -sqrt2 sin (x-pi/4)| =sqrt2|sin (x-pi/4)| <=sqrt2#. Answer link Related questions How do you find the x coordinates of the turning points of the function? How do you find the turning points of a cubic function? How many turning points can a cubic function have? How do you find the coordinates of the local extrema of the function? How do you find the local extrema of a function? How many local extrema can a cubic function have? How do I find the maximum and minimum values of the function #f(x) = x - 2 sin (x)# on the... If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum? How do you find the maximum of #f(x) = 2sin(x^2)#? How do you find a local minimum of a graph using the first derivative? See all questions in Identifying Turning Points (Local Extrema) for a Function Impact of this question 1821 views around the world You can reuse this answer Creative Commons License