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

Question

47245 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-16T14:15:03

#f_max=sqrt2#
#f_min=-sqrt2#

#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#