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

The extreme f(x)=(4x-3)/(x^2-4x+4) is Minimum(-1/2 ,-4/5)

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extreme f(x)=(4x-3)/(x^2-4x+4)

extreme points

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

Minimum (- \frac{1}{2}, - \frac{4}{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{4 x - 3}{x ^{2} - 4 x + 4} : - \frac{4}{5}

Minimum (- \frac{1}{2}, - \frac{4}{5})

What is the extreme f(x)=(4x-3)/(x^2-4x+4) ?
The extreme f(x)=(4x-3)/(x^2-4x+4) is Minimum(-1/2 ,-4/5)