What is the extreme f(x)= 1/8 (7x-1),x<= 3 ?

The extreme f(x)= 1/8 (7x-1),x<= 3 is Maximum(3, 5/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)= 1/8 (7x-1),x<= 3

extreme points

f (x) = \frac{1}{8} (7 x - 1), x \leq 3

Maximum (3, \frac{5}{2})

Solution steps

Find the critical points:

x = 3

Domain of

\frac{1}{8} (7 x - 1), x \leq 3 : x \leq 3

Find intervals:

Increasing

: - \infty < x < 3

Plug the extreme point

x = 3

into

\frac{1}{8} (7 x - 1) \Rightarrow y = \frac{5}{2}

Maximum (3, \frac{5}{2})

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

f (x, y) = e^{2 - x^{2} - y^{2}}

f (x) = - 12 x^{2} + 100

f (x) = 2 (x - 1) (x - 2) (x + 3)

C (x) = 30 x + 3300 I (x) - 10 x^{2} + 400 x

What is the extreme f(x)= 1/8 (7x-1),x<= 3 ?
The extreme f(x)= 1/8 (7x-1),x<= 3 is Maximum(3, 5/2)