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

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

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

extreme points

\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)

f (x) = \frac{1}{x + 6}

f (x) = 2 x^{2} - 4 x - 5

f (x) = lo g_{2} (x + 5) - 9

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

f (x) = - \frac{2 x}{x - 1}

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