What is the extreme f(x)=7sin(x)+7cos(x) ?

The extreme f(x)=7sin(x)+7cos(x) is Maximum(pi/4+2pin,7sqrt(2)),Minimum((5pi)/4+2pin,-7sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=7sin(x)+7cos(x)

extreme

Maximum (\frac{π}{4} + 2 πn, 72), Minimum (\frac{5 π}{4} + 2 πn, - 72)

Solution steps

Find the critical points:

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

Domain of

7 sin (x) + 7 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

7 sin (x) + 7 cos (x) \Rightarrow y = 72

Maximum (\frac{π}{4} + 2 πn, 72)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

7 sin (x) + 7 cos (x) \Rightarrow y = - 72

Minimum (\frac{5 π}{4} + 2 πn, - 72)

Maximum (\frac{π}{4} + 2 πn, 72), Minimum (\frac{5 π}{4} + 2 πn, - 72)

What is the extreme f(x)=7sin(x)+7cos(x) ?
The extreme f(x)=7sin(x)+7cos(x) is Maximum(pi/4+2pin,7sqrt(2)),Minimum((5pi)/4+2pin,-7sqrt(2))