What is the extreme f(x)=(x^2-x+1)/(x^2+1) ?

The extreme f(x)=(x^2-x+1)/(x^2+1) is Maximum(-1, 3/2),Minimum(1, 1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2-x+1)/(x^2+1)

extreme

Maximum (- 1, \frac{3}{2}), Minimum (1, \frac{1}{2})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{x ^{2} - x + 1}{x ^{2} + 1} : \frac{3}{2}

Maximum (- 1, \frac{3}{2})

Plug

x = 1

into

\frac{x ^{2} - x + 1}{x ^{2} + 1} : \frac{1}{2}

Minimum (1, \frac{1}{2})

Maximum (- 1, \frac{3}{2}), Minimum (1, \frac{1}{2})

What is the extreme f(x)=(x^2-x+1)/(x^2+1) ?
The extreme f(x)=(x^2-x+1)/(x^2+1) is Maximum(-1, 3/2),Minimum(1, 1/2)