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 1761 views around the world You can reuse this answer Creative Commons License