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

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

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

extreme points

y = \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)

\frac{1}{( x + 1 ) ^{2}}

f (x) = x (x^{2} - 12)

f (x, y) = (x + y^{2}) \cdot e^{2 x}

f (x, y) = 8 e^{(\frac{( - x ^{2} - y ^{2} )}{4})}

f (x, y) = ln (16 - x^{2} - 16 y^{2})

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