What is the extreme f(x)= x/(x^2-x+4),0<= x<= 6 ?

The extreme f(x)= x/(x^2-x+4),0<= x<= 6 is Minimum(0,0),Maximum(2, 1/3),Minimum(6, 3/17)

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extreme f(x)= x/(x^2-x+4),0<= x<= 6

extreme points

f (x) = \frac{x}{x ^{2} - x + 4}, 0 \leq x \leq 6

Minimum (0, 0), Maximum (2, \frac{1}{3}), Minimum (6, \frac{3}{17})

Solution steps

Find the critical points:

x = 0, x = 2, x = 6

Domain of

\frac{x}{x ^{2} - x + 4}, 0 \leq x \leq 6 : 0 \leq x \leq 6

Find intervals:

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 6

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 4} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 2

into

\frac{x}{x ^{2} - x + 4} \Rightarrow y = \frac{1}{3}

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

Plug the extreme point

x = 6

into

\frac{x}{x ^{2} - x + 4} \Rightarrow y = \frac{3}{17}

Minimum (6, \frac{3}{17})

Minimum (0, 0), Maximum (2, \frac{1}{3}), Minimum (6, \frac{3}{17})

f (x) = x^{3} - 6 x^{2} + 11

f (x, y) = 6 x - 4 y - x^{2} - 2 y^{2}

f (x, y) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

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

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

What is the extreme f(x)= x/(x^2-x+4),0<= x<= 6 ?
The extreme f(x)= x/(x^2-x+4),0<= x<= 6 is Minimum(0,0),Maximum(2, 1/3),Minimum(6, 3/17)