Find the absolute extrema of the given function f(x)= sinx+cosx on interval [0,2pi]?

1 Answer
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