What is the extreme f(x)= x/(x^2-x+9),0<= x<= 9 ?

The extreme f(x)= x/(x^2-x+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 1/5),Minimum(9, 1/9)

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extreme f(x)= x/(x^2-x+9),0<= x<= 9

extreme points

f (x) = \frac{x}{x ^{2} - x + 9}, 0 \leq x \leq 9

Minimum (0, 0), Maximum (3, \frac{1}{5}), Minimum (9, \frac{1}{9})

Solution steps

Find the critical points:

x = 0, x = 3, x = 9

Domain of

\frac{x}{x ^{2} - x + 9}, 0 \leq x \leq 9 : 0 \leq x \leq 9

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < 9

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 9} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 3

into

\frac{x}{x ^{2} - x + 9} \Rightarrow y = \frac{1}{5}

Maximum (3, \frac{1}{5})

Plug the extreme point

x = 9

into

\frac{x}{x ^{2} - x + 9} \Rightarrow y = \frac{1}{9}

Minimum (9, \frac{1}{9})

Minimum (0, 0), Maximum (3, \frac{1}{5}), Minimum (9, \frac{1}{9})

f (x) = e^{x^{2} - 3 x - 1}

f (x) = 3 x^{4} + 16 x^{3}

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

f (x) = (x - 1) e^{x}

\frac{x ^{2} + 1}{x ^{2} - 4}

What is the extreme f(x)= x/(x^2-x+9),0<= x<= 9 ?
The extreme f(x)= x/(x^2-x+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 1/5),Minimum(9, 1/9)