What is the extreme (x-3)/(x^2-18x+81) ?

The extreme (x-3)/(x^2-18x+81) is Minimum(-3,-1/24)

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extreme (x-3)/(x^2-18x+81)

extreme points

\frac{x - 3}{x ^{2} - 18 x + 81}

Minimum (- 3, - \frac{1}{24})

Solution steps

f^{'} (x) = - \frac{x + 3}{( x - 9 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 9,

Decreasing

: 9 < x < \infty

Plug

x = - 3

into

\frac{x - 3}{x ^{2} - 18 x + 81} : - \frac{1}{24}

Minimum (- 3, - \frac{1}{24})

x - \frac{54}{x}

f (x) = 2 x^{3} + 3 x^{2} - 120 x + 7

f (x) = 2 x^{3} + 3 x^{2} - 120 x + 1

f (x) = (\frac{4860}{x}) + 17 x + 752.774

f (x, y) = 2 y^{3} - 6 x y - x^{2}

What is the extreme (x-3)/(x^2-18x+81) ?
The extreme (x-3)/(x^2-18x+81) is Minimum(-3,-1/24)