What is the extreme f(x)=8sin(2x) ?

The extreme f(x)=8sin(2x) is Maximum(pi/4+pin,8),Minimum((3pi)/4+pin,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=8sin(2x)

extreme points

f (x) = 8 sin (2 x)

Maximum (\frac{π}{4} + πn, 8), Minimum (\frac{3 π}{4} + πn, - 8)

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

8 sin (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{4} + πn,

Decreasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Increasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

8 sin (2 x) \Rightarrow y = 8

Maximum (\frac{π}{4} + πn, 8)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

8 sin (2 x) \Rightarrow y = - 8

Minimum (\frac{3 π}{4} + πn, - 8)

Maximum (\frac{π}{4} + πn, 8), Minimum (\frac{3 π}{4} + πn, - 8)

f (x) = - 20 x^{2} - 30 y^{2} + 560 x + 360 y + 180

\frac{3 x + 5}{5 x ^{2} + 55 x + 120}

f (x) = x^{4} e^{- 7 x}

y = 8 + \frac{2}{x}

What is the extreme f(x)=8sin(2x) ?
The extreme f(x)=8sin(2x) is Maximum(pi/4+pin,8),Minimum((3pi)/4+pin,-8)