What are the extrema of #f(x)=-sinx-cosx# on the interval #[0,2pi]#?

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Answer 1 · 2015-11-08T15:49:47

Since #f(x)# is differentiable everywhere, simply find where #f'(x)=0#

#f'(x)=sin(x)-cos(x)=0#

Solve:

#sin(x)=cos(x)#

Now, either use the unit circle or sketch a graph of both functions to determine where they are equal:

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On the interval #[0,2pi]#, the two solutions are:

#x=pi/4# (minimum) or #(5pi)/4# (maximum)

hope that helps