Homework Soursefx sinxcosx 02pi from here I know it continues like this fx
f(x)= sinx-cosx [0,2pi]
from here I know it continues like this..
f\'(x)= cosx+sinx
f\'\'(x)=-sinx+cosx
sinx=cosx
x=pi/4, 5pi/4 <----this is the part I do not understand
Discuss the concavity of the graph of the function??
from here I know it continues like this..
f\'(x)= cosx+sinx
f\'\'(x)=-sinx+cosx
sinx=cosx
x=pi/4, 5pi/4 <----this is the part I do not understand
Discuss the concavity of the graph of the function??
Solution
f\'(x) = cosx + sinx f\'\'(x) = -sinx + cosx point of inflection f\'\'(x) = 0 sinx = cosx in [0 to 2pi] occurs at pi/4 and 5pi/4 (because at pi/4 both sin and cos are 1/sqrt(2) and at 5pi/4 both sin and cos are -1/sqrt(2)) so x = pi/4 and 5pi/4 are point of inflections![f(x)= sinx-cosx [0,2pi] from here I know it continues like this.. f\'(x)= cosx+sinx f\'\'(x)=-sinx+cosx sinx=cosx x=pi/4, 5pi/4 <----this is the part I do no](/WebImages/20/fx-sinxcosx-02pi-from-here-i-know-it-continues-like-this-fx-1044645-1761543212-0.webp "f(x)= sinx-cosx [0,2pi] from here I know it continues like this.. f\'(x)= cosx+sinx f\'\'(x)=-sinx+cosx sinx=cosx x=pi/4, 5pi/4 <----this is the part I do no")
