What are the extrema of #g(x) = 2 sin(2x - pi) + 4# on #[-pi/2,pi/2]#? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Ratnesh Bhosale May 27, 2018 #x=+-pi/4# for #x in[-pi/2,pi/2]# Explanation: #g(x) =2sin(2x-pi)+4# #g(x) =-2sin(2x)+4# For extrema of #g(x)#, #g'(x)=0# #g'(x) =-4cos(2x)# #g'(x)=0# #-4cos(2x)=0# #cos(2x)=0# #2x=+-pi/2# #x=+-pi/4# for #x in[-pi/2,pi/2]# 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 1572 views around the world You can reuse this answer Creative Commons License