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

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

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Answer 1 · 2016-04-09T10:49:47

Max $f (x) = \sqrt{2}$ $f(x)=sqrt2$ when
$x = \frac{5 π}{4} + k 2 π$ $x=(5pi)/4+k2pi$ or $x = - \frac{3 π}{4} + k 2 π$ $x=-(3pi)/4+k2pi$
Min $f (x) = - \sqrt{2}$ $f(x)=-sqrt2$ when
$x = \frac{π}{4} + k 2 π$ $x=pi/4+k2pi$

Explanation:

$f (x) = - sin x - cos x$ $f(x)=-sin x-cosx$
$f (x) = - (sin x - s i x (\frac{π}{2} - x))$ $f(x)=-(sinx-six(pi/2-x))$
$f (x) = - 2 sin (\frac{π}{4}) \cdot cos (x - \frac{π}{4})$ $f(x)=-2sin(pi/4)*cos(x-pi/4)$
$f (x) = - \sqrt{2} \cdot cos (x - \frac{π}{4})$ $f(x)=-sqrt2*cos(x-pi/4)$
max $f (x)$ $f(x)$ when $cos (x - \frac{π}{4}) = - 1$ $cos(x-pi/4)=-1$
Min $f (x)$ $f(x)$ when $cos (x - \frac{π}{4}) = - 1$ $cos(x-pi/4)=-1$

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

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