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

The extreme f(x)=(x^2-2x+1)/(x-4) is Maximum(1,0),Minimum(7,12)

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

extreme points

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

Maximum (1, 0), Minimum (7, 12)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 7}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 4,

Decreasing

: 4 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 2 x + 1}{x - 4} : 0

Maximum (1, 0)

Plug

x = 7

into

\frac{x ^{2} - 2 x + 1}{x - 4} : 12

Minimum (7, 12)

Maximum (1, 0), Minimum (7, 12)

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

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

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

f (x) = (x - 3) (x - 7)^{3} + 12, (1, 4)

f (x, y) = 2 x^{3} + y^{2} - 6 x y - 23

What is the extreme f(x)=(x^2-2x+1)/(x-4) ?
The extreme f(x)=(x^2-2x+1)/(x-4) is Maximum(1,0),Minimum(7,12)