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

The extreme f(x)=((x+1))/(x^2+x+1) is Minimum(-2,-1/3),Maximum(0,1)

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

extreme points

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

Minimum (- 2, - \frac{1}{3}), Maximum (0, 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

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

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

Plug

x = 0

into

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

Maximum (0, 1)

Minimum (- 2, - \frac{1}{3}), Maximum (0, 1)

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

\frac{ln ( x )}{x}

f (x) = x, 0 \leq x \leq 16

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

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 10

What is the extreme f(x)=((x+1))/(x^2+x+1) ?
The extreme f(x)=((x+1))/(x^2+x+1) is Minimum(-2,-1/3),Maximum(0,1)