What is the extreme y=(x^2-6x+9)/((x-1)^2) ?

The extreme y=(x^2-6x+9)/((x-1)^2) is Minimum(3,0)

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extreme y=(x^2-6x+9)/((x-1)^2)

extreme points

y = \frac{x ^{2} - 6 x + 9}{( x - 1 ) ^{2}}

Minimum (3, 0)

Solution steps

f^{'} (x) = \frac{4 ( x - 3 )}{( x - 1 ) ^{3}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 3

into

\frac{x ^{2} - 6 x + 9}{( x - 1 ) ^{2}} : 0

Minimum (3, 0)

f (x, y) = x^{2} + 4 y^{2} + 1

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

f (x) = \frac{x}{1 - x}

f (x) = 2 x + 4 y

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

What is the extreme y=(x^2-6x+9)/((x-1)^2) ?
The extreme y=(x^2-6x+9)/((x-1)^2) is Minimum(3,0)