What is the extreme f(x)=((x-3)^2)/(x-5) ?

The extreme f(x)=((x-3)^2)/(x-5) is Maximum(3,0),Minimum(7,8)

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extreme f(x)=((x-3)^2)/(x-5)

extreme points

f (x) = \frac{( x - 3 ) ^{2}}{x - 5}

Maximum (3, 0), Minimum (7, 8)

Solution steps

f^{'} (x) = \frac{x ^{2} - 10 x + 21}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 5,

Decreasing

: 5 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 3

into

\frac{( x - 3 ) ^{2}}{x - 5} : 0

Maximum (3, 0)

Plug

x = 7

into

\frac{( x - 3 ) ^{2}}{x - 5} : 8

Minimum (7, 8)

Maximum (3, 0), Minimum (7, 8)

What is the extreme f(x)=((x-3)^2)/(x-5) ?
The extreme f(x)=((x-3)^2)/(x-5) is Maximum(3,0),Minimum(7,8)