What is the extreme f(x)=((4x-18))/((x-9)^2) ?

The extreme f(x)=((4x-18))/((x-9)^2) is Minimum(0,-2/9)

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extreme f(x)=((4x-18))/((x-9)^2)

extreme points

f (x) = \frac{( 4 x - 18 )}{( x - 9 ) ^{2}}

Minimum (0, - \frac{2}{9})

Solution steps

f^{'} (x) = - \frac{4 x}{( x - 9 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 9,

Decreasing

: 9 < x < \infty

Plug

x = 0

into

\frac{4 x - 18}{( x - 9 ) ^{2}} : - \frac{2}{9}

Minimum (0, - \frac{2}{9})

f (x, y) = 11 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

y (x, t) = x (- 2 t + 2)

y = 2 x^{3} - 6 x

- \frac{6 x}{3 x ^{2} + 8}

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

What is the extreme f(x)=((4x-18))/((x-9)^2) ?
The extreme f(x)=((4x-18))/((x-9)^2) is Minimum(0,-2/9)