What is the extreme f(x)= x/(x^2-x+16),0<= x<= 18 ?

The extreme f(x)= x/(x^2-x+16),0<= x<= 18 is Minimum(0,0),Maximum(4, 1/7),Minimum(18, 9/161)

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extreme f(x)= x/(x^2-x+16),0<= x<= 18

extreme

Minimum (0, 0), Maximum (4, \frac{1}{7}), Minimum (18, \frac{9}{161})

Solution steps

Find the critical points:

x = 0, x = 4, x = 18

Domain of

\frac{x}{x ^{2} - x + 16}, 0 \leq x \leq 18 : 0 \leq x \leq 18

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 18

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 16} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 4

into

\frac{x}{x ^{2} - x + 16} \Rightarrow y = \frac{1}{7}

Maximum (4, \frac{1}{7})

Plug the extreme point

x = 18

into

\frac{x}{x ^{2} - x + 16} \Rightarrow y = \frac{9}{161}

Minimum (18, \frac{9}{161})

Minimum (0, 0), Maximum (4, \frac{1}{7}), Minimum (18, \frac{9}{161})

What is the extreme f(x)= x/(x^2-x+16),0<= x<= 18 ?
The extreme f(x)= x/(x^2-x+16),0<= x<= 18 is Minimum(0,0),Maximum(4, 1/7),Minimum(18, 9/161)