What is the extreme f(x)=x+9/x+2,1<= x<= 18 ?

The extreme f(x)=x+9/x+2,1<= x<= 18 is Maximum(1,12),Minimum(3,8),Maximum(18, 41/2)

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extreme f(x)=x+9/x+2,1<= x<= 18

extreme points

f (x) = x + \frac{9}{x} + 2, 1 \leq x \leq 18

Maximum (1, 12), Minimum (3, 8), Maximum (18, \frac{41}{2})

Solution steps

Find the critical points:

x = 1, x = 3, x = 18

Domain of

x + \frac{9}{x} + 2, 1 \leq x \leq 18 : 1 \leq x \leq 18

Find intervals:

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < 18

Plug the extreme point

x = 1

into

x + \frac{9}{x} + 2 \Rightarrow y = 12

Maximum (1, 12)

Plug the extreme point

x = 3

into

x + \frac{9}{x} + 2 \Rightarrow y = 8

Minimum (3, 8)

Plug the extreme point

x = 18

into

x + \frac{9}{x} + 2 \Rightarrow y = \frac{41}{2}

Maximum (18, \frac{41}{2})

Maximum (1, 12), Minimum (3, 8), Maximum (18, \frac{41}{2})

f (x) = 7 x^{4} - 6 x^{2} - 1

x^{2} - 2 x y + y^{3} - y

f (x) = 4 x + \frac{4}{x}

f (x) = 7 x^{4} - 6 x^{2} - 6

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

What is the extreme f(x)=x+9/x+2,1<= x<= 18 ?
The extreme f(x)=x+9/x+2,1<= x<= 18 is Maximum(1,12),Minimum(3,8),Maximum(18, 41/2)