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

The extreme f(x)=8sin(x)+8cos(x) is Maximum(pi/4+2pin,8sqrt(2)),Minimum((5pi)/4+2pin,-8sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=8sin(x)+8cos(x)

extreme points

f (x) = 8 sin (x) + 8 cos (x)

Maximum (\frac{π}{4} + 2 πn, 82), Minimum (\frac{5 π}{4} + 2 πn, - 82)

Solution steps

Find the critical points:

x = \frac{π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Domain of

8 sin (x) + 8 cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{4} + 2 πn,

Decreasing

: \frac{π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn,

Increasing

: \frac{5 π}{4} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{4} + 2 πn

into

8 sin (x) + 8 cos (x) \Rightarrow y = 82

Maximum (\frac{π}{4} + 2 πn, 82)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

8 sin (x) + 8 cos (x) \Rightarrow y = - 82

Minimum (\frac{5 π}{4} + 2 πn, - 82)

Maximum (\frac{π}{4} + 2 πn, 82), Minimum (\frac{5 π}{4} + 2 πn, - 82)

f (x) = 5 x + 5 sin (x), 0 \leq x \leq 2 π

f (x) = \frac{1}{2} x + 2, - 3 \leq x \leq 3

f (x) = \frac{x ^{2}}{x ^{2} + 27}

f (x) = 2 x^{2} + 4 x + 10

f (x) = 1 - x^{3} - 3 x - 2 x^{2} + 3 x^{4}

What is the extreme f(x)=8sin(x)+8cos(x) ?
The extreme f(x)=8sin(x)+8cos(x) is Maximum(pi/4+2pin,8sqrt(2)),Minimum((5pi)/4+2pin,-8sqrt(2))