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

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Answer 1 · 2015-11-08T15:49:47

Since $f (x)$ $f(x)$ 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

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