What is the extreme f(x)=(ln(x))/x ,1<= x<= 3 ?

The extreme f(x)=(ln(x))/x ,1<= x<= 3 is Minimum(1,0),Maximum(e, 1/e),Minimum(3,(ln(3))/3)

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extreme f(x)=(ln(x))/x ,1<= x<= 3

extreme points

f (x) = \frac{ln ( x )}{x}, 1 \leq x \leq 3

Minimum (1, 0), Maximum (e, \frac{1}{e}), Minimum (3, \frac{ln ( 3 )}{3})

Solution steps

Find the critical points:

x = 1, x = e, x = 3

Domain of

\frac{ln ( x )}{x}, 1 \leq x \leq 3 : 1 \leq x \leq 3

Find intervals:

Increasing

: 1 < x < e,

Decreasing

: e < x < 3

Plug the extreme point

x = 1

into

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

Minimum (1, 0)

Plug the extreme point

x = e

into

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

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

Plug the extreme point

x = 3

into

\frac{ln ( x )}{x} \Rightarrow y = \frac{ln ( 3 )}{3}

Minimum (3, \frac{ln ( 3 )}{3})

Minimum (1, 0), Maximum (e, \frac{1}{e}), Minimum (3, \frac{ln ( 3 )}{3})

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

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

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

f (x, y) = 10 - \frac{x ^{2}}{2} - \frac{y ^{2}}{2}

f (x, y) = x^{2} + y^{2} + x + y + x y

What is the extreme f(x)=(ln(x))/x ,1<= x<= 3 ?
The extreme f(x)=(ln(x))/x ,1<= x<= 3 is Minimum(1,0),Maximum(e, 1/e),Minimum(3,(ln(3))/3)