What is the extreme f(x)=(x^2-8x+37)/(x-6) ?

The extreme f(x)=(x^2-8x+37)/(x-6) is Maximum(1,-6),Minimum(11,14)

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extreme f(x)=(x^2-8x+37)/(x-6)

extreme points

f (x) = \frac{x ^{2} - 8 x + 37}{x - 6}

Maximum (1, - 6), Minimum (11, 14)

Solution steps

f^{'} (x) = \frac{x ^{2} - 12 x + 11}{( x - 6 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 6,

Decreasing

: 6 < x < 11,

Increasing

: 11 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 8 x + 37}{x - 6} : - 6

Maximum (1, - 6)

Plug

x = 11

into

\frac{x ^{2} - 8 x + 37}{x - 6} : 14

Minimum (11, 14)

Maximum (1, - 6), Minimum (11, 14)

f (x) = x^{4} + 8 x^{3} + 18 x^{2} + 2

89 e^{x^{4}} + 11

f (x y) = x^{2} + y^{2} + \frac{2}{x y}

f (x) = x^{3} - 12 x y + 8 y^{3}

f (x) = x 9 - x^{2}, - 1 \leq x \leq 3

What is the extreme f(x)=(x^2-8x+37)/(x-6) ?
The extreme f(x)=(x^2-8x+37)/(x-6) is Maximum(1,-6),Minimum(11,14)