What is the extreme f(x)=x+9/x ,0.2<= x<= 12 ?

The extreme f(x)=x+9/x ,0.2<= x<= 12 is Maximum(0.2,45.2),Minimum(3,6),Maximum(12, 51/4)

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extreme f(x)=x+9/x ,0.2<= x<= 12

extreme points

f (x) = x + \frac{9}{x}, 0.2 \leq x \leq 12

Maximum (0.2, 45.2), Minimum (3, 6), Maximum (12, \frac{51}{4})

Solution steps

Find the critical points:

x = 0.2, x = 3, x = 12

Domain of

x + \frac{9}{x}, 0.2 \leq x \leq 12 : 0.2 \leq x \leq 12

Find intervals:

Decreasing

: 0.2 < x < 3,

Increasing

: 3 < x < 12

Plug the extreme point

x = 0.2

into

x + \frac{9}{x} \Rightarrow y = 45.2

Maximum (0.2, 45.2)

Plug the extreme point

x = 3

into

x + \frac{9}{x} \Rightarrow y = 6

Minimum (3, 6)

Plug the extreme point

x = 12

into

x + \frac{9}{x} \Rightarrow y = \frac{51}{4}

Maximum (12, \frac{51}{4})

Maximum (0.2, 45.2), Minimum (3, 6), Maximum (12, \frac{51}{4})

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

f (x) = 5 - x + x^{2}

f (x, y) = \frac{- 2 x}{y} + e^{x} - 13

f (x) = - \frac{5 x ^{3}}{3} + 10 x^{2} - 15 x

f (x, y) = 2 x e^{- y}

What is the extreme f(x)=x+9/x ,0.2<= x<= 12 ?
The extreme f(x)=x+9/x ,0.2<= x<= 12 is Maximum(0.2,45.2),Minimum(3,6),Maximum(12, 51/4)