What is the extreme x/(x^2-x+25),0<= x<= 27 ?

The extreme x/(x^2-x+25),0<= x<= 27 is Minimum(0,0),Maximum(5, 1/9),Minimum(27, 27/727)

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extreme x/(x^2-x+25),0<= x<= 27

extreme points

\frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 27

Minimum (0, 0), Maximum (5, \frac{1}{9}), Minimum (27, \frac{27}{727})

Solution steps

Find the critical points:

x = 0, x = 5, x = 27

Domain of

\frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 27 : 0 \leq x \leq 27

Find intervals:

Increasing

: 0 < x < 5,

Decreasing

: 5 < x < 27

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 5

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = \frac{1}{9}

Maximum (5, \frac{1}{9})

Plug the extreme point

x = 27

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = \frac{27}{727}

Minimum (27, \frac{27}{727})

Minimum (0, 0), Maximum (5, \frac{1}{9}), Minimum (27, \frac{27}{727})

f (x) = 5 (x - 9)^{3}

f (x) = x^{2} - x y + y^{2} - 1 x + 1 y

x + f (x) = x^{2} + 13

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

f (x) = 3 x - 1

What is the extreme x/(x^2-x+25),0<= x<= 27 ?
The extreme x/(x^2-x+25),0<= x<= 27 is Minimum(0,0),Maximum(5, 1/9),Minimum(27, 27/727)