What is the extreme f(x)=((x^2-5x+4))/(x-5) ?

The extreme f(x)=((x^2-5x+4))/(x-5) is Maximum(3,1),Minimum(7,9)

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extreme f(x)=((x^2-5x+4))/(x-5)

extreme

Maximum (3, 1), Minimum (7, 9)

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 ^{2} - 5 x + 4}{x - 5} : 1

Maximum (3, 1)

Plug

x = 7

into

\frac{x ^{2} - 5 x + 4}{x - 5} : 9

Minimum (7, 9)

Maximum (3, 1), Minimum (7, 9)

What is the extreme f(x)=((x^2-5x+4))/(x-5) ?
The extreme f(x)=((x^2-5x+4))/(x-5) is Maximum(3,1),Minimum(7,9)