What is the extreme h(x)=(χ^2+4x-32)/(x^2-8x+16) ?

The extreme h(x)=(χ^2+4x-32)/(x^2-8x+16) is Saddle(12,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme h(x)=(χ^2+4x-32)/(x^2-8x+16)

extreme points

h (x) = \frac{χ ^{2} + 4 x - 32}{x ^{2} - 8 x + 16}

Saddle (12, 0)

Solution steps

Find the critical points:

(12, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2}{x ^{2} - 8 x + 16}

D (x, χ) = \frac{- 256 + 16 x - 4 χ ^{2}}{( x - 4 ) ^{6}}

Check the sign of at each critical point

Check critical point

(12, 0) :

Saddle

Saddle (12, 0)

f (x) = 2 x^{3} 6 x^{2} - 18 x + 1

f (x) = 2 x^{3} - y^{2} - 6 x + 9 y

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

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

f (x, y) = 2 x^{2} + x y + y^{2} - x - 3 y

What is the extreme h(x)=(χ^2+4x-32)/(x^2-8x+16) ?
The extreme h(x)=(χ^2+4x-32)/(x^2-8x+16) is Saddle(12,0)