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

The extreme f(x)=((x-2)^2)/(x-4) is Maximum(2,0),Minimum(6,8)

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

extreme points

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

Maximum (2, 0), Minimum (6, 8)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Decreasing

: 4 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

\frac{( x - 2 ) ^{2}}{x - 4} : 0

Maximum (2, 0)

Plug

x = 6

into

\frac{( x - 2 ) ^{2}}{x - 4} : 8

Minimum (6, 8)

Maximum (2, 0), Minimum (6, 8)

(3.15) (13.3)

f (x) = \frac{x ^{2}}{2}

f (x) = \frac{8}{9} cos (\frac{π x}{3})

x^{2} + 5 x + 6

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

What is the extreme f(x)=((x-2)^2)/(x-4) ?
The extreme f(x)=((x-2)^2)/(x-4) is Maximum(2,0),Minimum(6,8)