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

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

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Answer 1 · 2016-04-09T10:49:47

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

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