What is the extreme f(x)=((x^2-11x+82))/(x-9) ?

The extreme f(x)=((x^2-11x+82))/(x-9) is Maximum(1,-9),Minimum(17,23)

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extreme f(x)=((x^2-11x+82))/(x-9)

extreme

Maximum (1, - 9), Minimum (17, 23)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x + 17}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 9,

Decreasing

: 9 < x < 17,

Increasing

: 17 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 11 x + 82}{x - 9} : - 9

Maximum (1, - 9)

Plug

x = 17

into

\frac{x ^{2} - 11 x + 82}{x - 9} : 23

Minimum (17, 23)

Maximum (1, - 9), Minimum (17, 23)

What is the extreme f(x)=((x^2-11x+82))/(x-9) ?
The extreme f(x)=((x^2-11x+82))/(x-9) is Maximum(1,-9),Minimum(17,23)