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

The extreme f(x)=(x^2-2x+4)/(x-2) is Maximum(0,-2),Minimum(4,6)

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

extreme points

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

Maximum (0, - 2), Minimum (4, 6)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

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

Maximum (0, - 2)

Plug

x = 4

into

\frac{x ^{2} - 2 x + 4}{x - 2} : 6

Minimum (4, 6)

Maximum (0, - 2), Minimum (4, 6)

What is the extreme f(x)=(x^2-2x+4)/(x-2) ?
The extreme f(x)=(x^2-2x+4)/(x-2) is Maximum(0,-2),Minimum(4,6)