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

The extreme f(x)=(x^2-4x+4)/(x-9) is Maximum(2,0),Minimum(16,28)

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

extreme points

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

Maximum (2, 0), Minimum (16, 28)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x + 32}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 9,

Decreasing

: 9 < x < 16,

Increasing

: 16 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 4 x + 4}{x - 9} : 0

Maximum (2, 0)

Plug

x = 16

into

\frac{x ^{2} - 4 x + 4}{x - 9} : 28

Minimum (16, 28)

Maximum (2, 0), Minimum (16, 28)

f (x) = \frac{2 x - 1}{x - 7}

- 2 cos (x)

f (x) = \frac{x ^{2} - 25}{x ^{2} - 6 x + 5}

\frac{1}{x - 10}

\frac{2 x + 1}{x - 3}

What is the extreme f(x)=(x^2-4x+4)/(x-9) ?
The extreme f(x)=(x^2-4x+4)/(x-9) is Maximum(2,0),Minimum(16,28)