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

The extreme f(x)=(x^2+1)/(x^2-4x+4) is Minimum(-1/2 , 1/5)

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extreme f(x)=(x^2+1)/(x^2-4x+4)

extreme points

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

Minimum (- \frac{1}{2}, \frac{1}{5})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - \frac{1}{2}

into

\frac{x ^{2} + 1}{x ^{2} - 4 x + 4} : \frac{1}{5}

Minimum (- \frac{1}{2}, \frac{1}{5})

f (x, y) = (x - y) (25 - x y)

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

f (x) = x y + \frac{3}{x} + \frac{9}{y}

f (x) = - 8 x^{2} + 8 x

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

What is the extreme f(x)=(x^2+1)/(x^2-4x+4) ?
The extreme f(x)=(x^2+1)/(x^2-4x+4) is Minimum(-1/2 , 1/5)