What is the extreme f(x)=(x^2-11x+54)/(x-9) ?

The extreme f(x)=(x^2-11x+54)/(x-9) is Maximum(3,-5),Minimum(15,19)

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extreme f(x)=(x^2-11x+54)/(x-9)

extreme points

f (x) = \frac{x ^{2} - 11 x + 54}{x - 9}

Maximum (3, - 5), Minimum (15, 19)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x + 45}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 9,

Decreasing

: 9 < x < 15,

Increasing

: 15 < x < \infty

Plug

x = 3

into

\frac{x ^{2} - 11 x + 54}{x - 9} : - 5

Maximum (3, - 5)

Plug

x = 15

into

\frac{x ^{2} - 11 x + 54}{x - 9} : 19

Minimum (15, 19)

Maximum (3, - 5), Minimum (15, 19)

2 x y

f (x) = \frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9

f (x) = 2 x + ln (x)

f (x) = - 2 x^{3} - 3 x^{2} + x + 1

f (x) = 2 θ - 3 sin (θ)

What is the extreme f(x)=(x^2-11x+54)/(x-9) ?
The extreme f(x)=(x^2-11x+54)/(x-9) is Maximum(3,-5),Minimum(15,19)