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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

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

Plug

x = 1

into

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

Maximum (1, 3)

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

f (x) = 6 - 8 x, - 8 \leq x \leq 6

2 x^{5} - 15 x^{4} + 30 x^{3} - 6

f (x, y) = 1 + x^{2} + y^{2}

p (a, c) = a + 8 + c

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

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