How do you find the local max and min for f(x)=-sinx-cosx?

1 Answer
Dang Thi ThanhHoa
Apr 9, 2016

Max f(x)=sqrt2 when
x=(5pi)/4+k2pi or x=-(3pi)/4+k2pi
Min f(x)=-sqrt2 when
x=pi/4+k2pi

Explanation:

f(x)=-sin x-cosx
f(x)=-(sinx-six(pi/2-x))
f(x)=-2sin(pi/4)*cos(x-pi/4)
f(x)=-sqrt2*cos(x-pi/4)
max f(x) when cos(x-pi/4)=-1
Min f(x) when cos(x-pi/4)=-1

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