What is the extreme x/(x^2-x+16),0<= x<= 12 ?

The extreme x/(x^2-x+16),0<= x<= 12 is Minimum(0,0),Maximum(4, 1/7),Minimum(12, 3/37)

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extreme x/(x^2-x+16),0<= x<= 12

extreme points

\frac{x}{x ^{2} - x + 16}, 0 \leq x \leq 12

Minimum (0, 0), Maximum (4, \frac{1}{7}), Minimum (12, \frac{3}{37})

Solution steps

Find the critical points:

x = 0, x = 4, x = 12

Domain of

\frac{x}{x ^{2} - x + 16}, 0 \leq x \leq 12 : 0 \leq x \leq 12

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 12

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 = 12

into

\frac{x}{x ^{2} - x + 16} \Rightarrow y = \frac{3}{37}

Minimum (12, \frac{3}{37})

Minimum (0, 0), Maximum (4, \frac{1}{7}), Minimum (12, \frac{3}{37})

f (x) = (\frac{( x ^{2} - 4 )}{( x ^{2} + 4 )})

f (x) = \frac{2}{3} x^{3} - 6 x^{2} + 16 x

f (x, y) = x^{2} + 3 x y + 4 y^{2} - 6 x + 2 y

t 4 - t

f (x) = \frac{1}{x ^{3} - x}

What is the extreme x/(x^2-x+16),0<= x<= 12 ?
The extreme x/(x^2-x+16),0<= x<= 12 is Minimum(0,0),Maximum(4, 1/7),Minimum(12, 3/37)