Question

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

Answer 1

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`