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

The extreme f(x)=(x^2-x-2)/(x^2-6x+9) is Minimum(7/5 ,-9/16)

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-x-2)/(x^2-6x+9)

extreme points

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

Minimum (\frac{7}{5}, - \frac{9}{16})

Solution steps

f^{'} (x) = - \frac{5 x - 7}{( x - 3 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < \frac{7}{5},

Increasing

: \frac{7}{5} < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = \frac{7}{5}

into

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

Minimum (\frac{7}{5}, - \frac{9}{16})

f (x, y) = x y + \frac{8}{x} + \frac{8}{y}

f (x, y) = x y - 7 x y + 200 x

f (x, y) = x y - 5 x - 5 y - x^{2} - y^{2}

f (x) = \frac{x - 1}{x ^{2}}

f (x, y) = - 5 x^{2} + 4 x y - y^{2} + 16 x + 10

What is the extreme f(x)=(x^2-x-2)/(x^2-6x+9) ?
The extreme f(x)=(x^2-x-2)/(x^2-6x+9) is Minimum(7/5 ,-9/16)