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

The extreme f(x)=(x^2-2x+1)/(x-9) is Maximum(1,0),Minimum(17,32)

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

extreme points

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

Maximum (1, 0), Minimum (17, 32)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x + 17}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 9,

Decreasing

: 9 < x < 17,

Increasing

: 17 < x < \infty

Plug

x = 1

into

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

Maximum (1, 0)

Plug

x = 17

into

\frac{x ^{2} - 2 x + 1}{x - 9} : 32

Minimum (17, 32)

Maximum (1, 0), Minimum (17, 32)

f (x) = (x + 3)^{4} + 3

e^{2 x} (x^{2} - 2)

\frac{2 x}{1 - x ^{2}}

z = 4 x + 5 y

f (x) = \frac{( x ^{2} - 7 )}{x + 4}

What is the extreme f(x)=(x^2-2x+1)/(x-9) ?
The extreme f(x)=(x^2-2x+1)/(x-9) is Maximum(1,0),Minimum(17,32)