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 points

f (x) = \frac{x ^{2} - 11 x + 67}{x - 9}

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)

f (x) = - x^{3} + 3 x + 7

f (x) = x ln (9 x)

\frac{3 - 14 x - 5 x ^{2}}{3 + 7 x + 2 x ^{2}}

f (x) = x^{\frac{2}{3}} (7 x + 5)

f (x, y) = x^{3} - 9 x y + y^{3} - 2

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)