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

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

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

extreme points

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

Minimum (\frac{3}{2}, \frac{9}{2} - 9 (ln (3) - \frac{1}{2} ln (2)))

Solution steps

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

Find intervals:

Decreasing

: 0 < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = \frac{3}{2}

into

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

Minimum (\frac{3}{2}, \frac{9}{2} - 9 (ln (3) - \frac{1}{2} ln (2)))

f (x) = (\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3})

f (x) = - 5 x^{3} + 45 x

f (x) = - 0.020724 x^{2} + 0.3692 x - 0.6287

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

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

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