Find the absolute extrema of the given function f(x)= sinx+cosx on interval [0,2pi]? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Sasha P. Oct 16, 2015 f_max=sqrt2 f_min=-sqrt2 Explanation: f'(x)=cosx-sinx f'(x)=0 <=> cosx-sinx=0 cosx/cosx-sinx/cosx=0 ^^ cosx !=0 tanx=1 ^^ cosx !=0 x=pi/4+kpi ^^ x != pi/2+mpi x=pi/4+kpi On [0,2pi] : x=pi/4 vv x=(5pi)/4 f_max=f(pi/4) = sin(pi/4)+cos(pi/4) = sqrt2/2+sqrt2/2=sqrt2 f_min=f((5pi)/4) = sin((5pi)/4)+cos((5pi)/4) = -sqrt2/2-sqrt2/2=-sqrt2 f_max=sqrt2 f_min=-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 47358 views around the world You can reuse this answer Creative Commons License