What is the extreme f(x)=(x^2-9x+39)/(x-7) ?

The extreme f(x)=(x^2-9x+39)/(x-7) is Maximum(2,-5),Minimum(12,15)

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extreme f(x)=(x^2-9x+39)/(x-7)

extreme points

f (x) = \frac{x ^{2} - 9 x + 39}{x - 7}

Maximum (2, - 5), Minimum (12, 15)

Solution steps

f^{'} (x) = \frac{x ^{2} - 14 x + 24}{( x - 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 7,

Decreasing

: 7 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 9 x + 39}{x - 7} : - 5

Maximum (2, - 5)

Plug

x = 12

into

\frac{x ^{2} - 9 x + 39}{x - 7} : 15

Minimum (12, 15)

Maximum (2, - 5), Minimum (12, 15)

4 x^{2} - 8 x + 16

x + \frac{3}{x}

f (x, y) = 6 x^{2} + y^{3} - 12 x y + 2

f (x) = x^{2} - 10

- x^{3} + 9 x^{2}

What is the extreme f(x)=(x^2-9x+39)/(x-7) ?
The extreme f(x)=(x^2-9x+39)/(x-7) is Maximum(2,-5),Minimum(12,15)