What is the extreme f(x)=(x^2-11x+67)/(x-9) ?

The extreme f(x)=(x^2-11x+67)/(x-9) is Maximum(2,-7),Minimum(16,21)

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extreme f(x)=(x^2-11x+67)/(x-9)

extreme

Maximum (2, - 7), Minimum (16, 21)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x + 32}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 9,

Decreasing

: 9 < x < 16,

Increasing

: 16 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 11 x + 67}{x - 9} : - 7

Maximum (2, - 7)

Plug

x = 16

into

\frac{x ^{2} - 11 x + 67}{x - 9} : 21

Minimum (16, 21)

Maximum (2, - 7), Minimum (16, 21)

What is the extreme f(x)=(x^2-11x+67)/(x-9) ?
The extreme f(x)=(x^2-11x+67)/(x-9) is Maximum(2,-7),Minimum(16,21)