What is the extreme f(x)=(ln(x))/(3x),1<= x<= 4 ?

The extreme f(x)=(ln(x))/(3x),1<= x<= 4 is Minimum(1,0),Maximum(e, 1/(3e)),Minimum(4,(ln(2))/6)

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extreme f(x)=(ln(x))/(3x),1<= x<= 4

extreme points

f (x) = \frac{ln ( x )}{3 x}, 1 \leq x \leq 4

Minimum (1, 0), Maximum (e, \frac{1}{3 e}), Minimum (4, \frac{ln ( 2 )}{6})

Solution steps

Find the critical points:

x = 1, x = e, x = 4

Domain of

\frac{ln ( x )}{3 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 )}{3 x} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = e

into

\frac{ln ( x )}{3 x} \Rightarrow y = \frac{1}{3 e}

Maximum (e, \frac{1}{3 e})

Plug the extreme point

x = 4

into

\frac{ln ( x )}{3 x} \Rightarrow y = \frac{ln ( 2 )}{6}

Minimum (4, \frac{ln ( 2 )}{6})

Minimum (1, 0), Maximum (e, \frac{1}{3 e}), Minimum (4, \frac{ln ( 2 )}{6})

f (x) = (1 - 2 x - x^{2} + 4 y^{2} - 2 y^{2})

3 + 6 x^{2} - 4 x^{3}

f (x, y) = x^{2} + y^{2} + 8 x - 4 y + 7

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

F (x, y) = 15 x^{2} - 22 x y + 24 x + 8 y^{2} - 16 y

What is the extreme f(x)=(ln(x))/(3x),1<= x<= 4 ?
The extreme f(x)=(ln(x))/(3x),1<= x<= 4 is Minimum(1,0),Maximum(e, 1/(3e)),Minimum(4,(ln(2))/6)