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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme x+cos(2x),0<= x<= pi

extreme points

x + cos (2 x), 0 \leq x \leq π

Minimum (0, 1), Maximum (\frac{π}{12}, \frac{π}{12} + \frac{3}{2}), Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2}), Maximum (π, π + 1)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{12}, x = \frac{5 π}{12}, x = π

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{12},

Decreasing

: \frac{π}{12} < x < \frac{5 π}{12},

Increasing

: \frac{5 π}{12} < x < π

Plug the extreme point

x = 0

into

x + cos (2 x) \Rightarrow y = 1

Minimum (0, 1)

Plug the extreme point

x = \frac{π}{12}

into

x + cos (2 x) \Rightarrow y = \frac{π}{12} + \frac{3}{2}

Maximum (\frac{π}{12}, \frac{π}{12} + \frac{3}{2})

Plug the extreme point

x = \frac{5 π}{12}

into

x + cos (2 x) \Rightarrow y = \frac{5 π}{12} - \frac{3}{2}

Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2})

Plug the extreme point

x = π

into

x + cos (2 x) \Rightarrow y = π + 1

Maximum (π, π + 1)

Minimum (0, 1), Maximum (\frac{π}{12}, \frac{π}{12} + \frac{3}{2}), Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2}), Maximum (π, π + 1)

f (x) = \frac{x ^{2}}{x ^{2} + 192}

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

f (x, y) = 3 x^{2} + y^{2} - x y - 11 x

f (t) = (t) + 2 u (t) - 3 e^{- 2 t} u (t)

f (x) = (x - 6) e^{x}