What is the extreme f(x)=(x^2-6x+9)/(x-7) ?

The extreme f(x)=(x^2-6x+9)/(x-7) is Maximum(3,0),Minimum(11,16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x^2-6x+9)/(x-7)

extreme points

f (x) = \frac{x ^{2} - 6 x + 9}{x - 7}

Maximum (3, 0), Minimum (11, 16)

Solution steps

f^{'} (x) = \frac{x ^{2} - 14 x + 33}{( x - 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 7,

Decreasing

: 7 < x < 11,

Increasing

: 11 < x < \infty

Plug

x = 3

into

\frac{x ^{2} - 6 x + 9}{x - 7} : 0

Maximum (3, 0)

Plug

x = 11

into

\frac{x ^{2} - 6 x + 9}{x - 7} : 16

Minimum (11, 16)

Maximum (3, 0), Minimum (11, 16)

f (x) = x 21 - x

f (x) = - x^{2} - 25

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

f (x) = x^{4} - x^{3} - 7 x^{2} + x + 6

- x^{2} - x + 3

What is the extreme f(x)=(x^2-6x+9)/(x-7) ?
The extreme f(x)=(x^2-6x+9)/(x-7) is Maximum(3,0),Minimum(11,16)