What is the extreme f(x)= 9/2 x^2-ln(x) ?

The extreme f(x)= 9/2 x^2-ln(x) is Minimum(1/3 , 1/2+ln(3))

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extreme f(x)= 9/2 x^2-ln(x)

extreme points

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

Minimum (\frac{1}{3}, \frac{1}{2} + ln (3))

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{1}{3}

into

\frac{9}{2} x^{2} - ln (x) : \frac{1}{2} + ln (3)

Minimum (\frac{1}{3}, \frac{1}{2} + ln (3))

f (x) = 3 x + 1

f (x) = \frac{5 ( x - 6 ) ^{2}}{x - 1}

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

f (x) = x^{3} - x^{2} - 8 x + 12

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

What is the extreme f(x)= 9/2 x^2-ln(x) ?
The extreme f(x)= 9/2 x^2-ln(x) is Minimum(1/3 , 1/2+ln(3))