What is the extreme f(x)=(ln(x))/(9x),1<= x<= 4 ?

The extreme f(x)=(ln(x))/(9x),1<= x<= 4 is Minimum(1,0),Maximum(e, 1/(9e)),Minimum(4,(ln(2))/(18))

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extreme f(x)=(ln(x))/(9x),1<= x<= 4

extreme points

f (x) = \frac{ln ( x )}{9 x}, 1 \leq x \leq 4

Minimum (1, 0), Maximum (e, \frac{1}{9 e}), Minimum (4, \frac{ln ( 2 )}{18})

Solution steps

Find the critical points:

x = 1, x = e, x = 4

Domain of

\frac{ln ( x )}{9 x}, 1 \leq x \leq 4 : 1 \leq x \leq 4

Find intervals:

Increasing

: 1 < x < e,

Decreasing

: e < x < 4

Plug the extreme point

x = 1

into

\frac{ln ( x )}{9 x} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = e

into

\frac{ln ( x )}{9 x} \Rightarrow y = \frac{1}{9 e}

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

Plug the extreme point

x = 4

into

\frac{ln ( x )}{9 x} \Rightarrow y = \frac{ln ( 2 )}{18}

Minimum (4, \frac{ln ( 2 )}{18})

Minimum (1, 0), Maximum (e, \frac{1}{9 e}), Minimum (4, \frac{ln ( 2 )}{18})

f (x) = \frac{x ^{2} - 2 x + 1}{x - 9}

f (x) = (x + 3)^{4} + 3

e^{2 x} (x^{2} - 2)

\frac{2 x}{1 - x ^{2}}

z = 4 x + 5 y

What is the extreme f(x)=(ln(x))/(9x),1<= x<= 4 ?
The extreme f(x)=(ln(x))/(9x),1<= x<= 4 is Minimum(1,0),Maximum(e, 1/(9e)),Minimum(4,(ln(2))/(18))