What is the extreme f(x)=-sin(x)-4,(0,2pi) ?

The extreme f(x)=-sin(x)-4,(0,2pi) is Minimum(pi/2 ,-5),Maximum((3pi)/2 ,-3)

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extreme f(x)=-sin(x)-4,(0,2pi)

extreme points

f (x) = - sin (x) - 4, (0, 2 π)

Minimum (\frac{π}{2}, - 5), Maximum (\frac{3 π}{2}, - 3)

Solution steps

Find the critical points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Domain of

- sin (x) - 4, (0, 2 π) : 0 < x < 2 π

Find intervals:

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < \frac{3 π}{2},

Decreasing

: \frac{3 π}{2} < x < 2 π

Plug the extreme point

x = \frac{π}{2}

into

- sin (x) - 4 \Rightarrow y = - 5

Minimum (\frac{π}{2}, - 5)

Plug the extreme point

x = \frac{3 π}{2}

into

- sin (x) - 4 \Rightarrow y = - 3

Maximum (\frac{3 π}{2}, - 3)

Minimum (\frac{π}{2}, - 5), Maximum (\frac{3 π}{2}, - 3)

f (x) = \frac{x + 5}{x ^{2} - 4}

f (x) = 6 + 9 x^{2} - 6 x^{3}

f (x) = 14 x + 17 x^{\frac{14}{17}}

f (x, y) = 2 x y - 6 y^{2} + x^{2} y - 3 y + 7

f (x) = 2 x^{4} - x^{2} + 5

What is the extreme f(x)=-sin(x)-4,(0,2pi) ?
The extreme f(x)=-sin(x)-4,(0,2pi) is Minimum(pi/2 ,-5),Maximum((3pi)/2 ,-3)