What is the extreme f(x)=12x-2ln(x),x>0 ?

The extreme f(x)=12x-2ln(x),x>0 is Minimum(1/6 ,2+2ln(6))

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extreme f(x)=12x-2ln(x),x>0

extreme points

f (x) = 12 x - 2 ln (x), x > 0

Minimum (\frac{1}{6}, 2 + 2 ln (6))

Solution steps

Find the critical points:

x = \frac{1}{6}

Domain of

12 x - 2 ln (x), x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{6},

Increasing

: \frac{1}{6} < x < \infty

Plug the extreme point

x = \frac{1}{6}

into

12 x - 2 ln (x) \Rightarrow y = 2 + 2 ln (6)

Minimum (\frac{1}{6}, 2 + 2 ln (6))

(x^{2} - 9)^{6}

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

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

f (x) = x [n] = 5 sin (2 n)

f (x) = \frac{x}{x - 3}

What is the extreme f(x)=12x-2ln(x),x>0 ?
The extreme f(x)=12x-2ln(x),x>0 is Minimum(1/6 ,2+2ln(6))