What is the extreme f(x)=x-2cos(x),0<= x<= 2pi ?

The extreme f(x)=x-2cos(x),0<= x<= 2pi is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x-2cos(x),0<= x<= 2pi

extreme points

f (x) = x - 2 cos (x), 0 \leq x \leq 2 π

Minimum (0, - 2), Maximum (\frac{7 π}{6}, \frac{7 π}{6} + 3), Minimum (\frac{11 π}{6}, \frac{11 π}{6} - 3), Maximum (2 π, 2 π - 2)

Solution steps

Find the critical points:

x = 0, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 2 π

Domain of

x - 2 cos (x), 0 \leq x \leq 2 π : 0 \leq x \leq 2 π

Find intervals:

Increasing

: 0 < x < \frac{7 π}{6},

Decreasing

: \frac{7 π}{6} < x < \frac{11 π}{6},

Increasing

: \frac{11 π}{6} < x < 2 π

Plug the extreme point

x = 0

into

x - 2 cos (x) \Rightarrow y = - 2

Minimum (0, - 2)

Plug the extreme point

x = \frac{7 π}{6}

into

x - 2 cos (x) \Rightarrow y = \frac{7 π}{6} + 3

Maximum (\frac{7 π}{6}, \frac{7 π}{6} + 3)

Plug the extreme point

x = \frac{11 π}{6}

into

x - 2 cos (x) \Rightarrow y = \frac{11 π}{6} - 3

Minimum (\frac{11 π}{6}, \frac{11 π}{6} - 3)

Plug the extreme point

x = 2 π

into

x - 2 cos (x) \Rightarrow y = 2 π - 2

Maximum (2 π, 2 π - 2)

Minimum (0, - 2), Maximum (\frac{7 π}{6}, \frac{7 π}{6} + 3), Minimum (\frac{11 π}{6}, \frac{11 π}{6} - 3), Maximum (2 π, 2 π - 2)

f (x) = x^{4} - 20 x^{3} + 28 x^{2} - 4

1448.2 x^{2} - 81667 x + 1000000

f (x, y) = (x - 8)^{2} + (y - 5)^{2} + 16

f (x) = x e^{3 x^{2}}

f (x, y) = 2 x^{2} + y^{4} - 16 x y

What is the extreme f(x)=x-2cos(x),0<= x<= 2pi ?
The extreme f(x)=x-2cos(x),0<= x<= 2pi is