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

The extreme f(x)= x/(x^2-x+1),0<= x<= 3 is Minimum(0,0),Maximum(1,1),Minimum(3, 3/7)

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extreme f(x)= x/(x^2-x+1),0<= x<= 3

extreme points

f (x) = \frac{x}{x ^{2} - x + 1}, 0 \leq x \leq 3

Minimum (0, 0), Maximum (1, 1), Minimum (3, \frac{3}{7})

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

\frac{x}{x ^{2} - x + 1}, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 1} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

\frac{x}{x ^{2} - x + 1} \Rightarrow y = 1

Maximum (1, 1)

Plug the extreme point

x = 3

into

\frac{x}{x ^{2} - x + 1} \Rightarrow y = \frac{3}{7}

Minimum (3, \frac{3}{7})

Minimum (0, 0), Maximum (1, 1), Minimum (3, \frac{3}{7})

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

F (I, J) = 25 I N + 12 J N

(x - x^{2})^{2}, - 1 \leq x \leq 1

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

f (x, y) = 4 x y

What is the extreme f(x)= x/(x^2-x+1),0<= x<= 3 ?
The extreme f(x)= x/(x^2-x+1),0<= x<= 3 is Minimum(0,0),Maximum(1,1),Minimum(3, 3/7)