What is the extreme f(x)=3x-ln(3x),(0,4) ?

The extreme f(x)=3x-ln(3x),(0,4) is Minimum(1/3 ,1)

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extreme f(x)=3x-ln(3x),(0,4)

extreme points

f (x) = 3 x - ln (3 x), (0, 4)

Minimum (\frac{1}{3}, 1)

Solution steps

Find the critical points:

x = \frac{1}{3}

Domain of

3 x - ln (3 x), (0, 4) : 0 < x < 4

Find intervals:

Decreasing

: 0 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < 4

Plug the extreme point

x = \frac{1}{3}

into

3 x - ln (3 x) \Rightarrow y = 1

Minimum (\frac{1}{3}, 1)

2 x^{3} - 9 x

f (x) = (2 x^{3} - 3 x^{2} - 12 x + 1)

f (1) = x \cdot e^{y}

x^{3} + 10 x^{2} + x + 1

f (x, y) = x^{2} + y^{2} + 12 x - 14 y

What is the extreme f(x)=3x-ln(3x),(0,4) ?
The extreme f(x)=3x-ln(3x),(0,4) is Minimum(1/3 ,1)