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

The extreme f(x)=(4x-2)/(x^2-4x+4) is Minimum(-1,-2/3)

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extreme f(x)=(4x-2)/(x^2-4x+4)

extreme points

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

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

Solution steps

f^{'} (x) = - \frac{4 ( x + 1 )}{( x - 2 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 1

into

\frac{4 x - 2}{x ^{2} - 4 x + 4} : - \frac{2}{3}

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

f (x) = ln (x^{2} + 7 x + 14)

f (x, y) = y^{3} + x^{2} - 6 x y + 3 x + 6 y - 7

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

f (x) = x^{3} - 5 x^{2} + 7 x

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

What is the extreme f(x)=(4x-2)/(x^2-4x+4) ?
The extreme f(x)=(4x-2)/(x^2-4x+4) is Minimum(-1,-2/3)