What is the extreme f(x)=(x^2-9x+50)/(x-7) ?

The extreme f(x)=(x^2-9x+50)/(x-7) is Maximum(1,-7),Minimum(13,17)

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extreme f(x)=(x^2-9x+50)/(x-7)

extreme points

f (x) = \frac{x ^{2} - 9 x + 50}{x - 7}

Maximum (1, - 7), Minimum (13, 17)

Solution steps

f^{'} (x) = \frac{x ^{2} - 14 x + 13}{( x - 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 7,

Decreasing

: 7 < x < 13,

Increasing

: 13 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 9 x + 50}{x - 7} : - 7

Maximum (1, - 7)

Plug

x = 13

into

\frac{x ^{2} - 9 x + 50}{x - 7} : 17

Minimum (13, 17)

Maximum (1, - 7), Minimum (13, 17)

- \frac{4}{x - 7}

f (x, y) = - 4 x^{2} - 2 y^{2} - 8 x + 12 y + 5

f (x, y) = \frac{4}{x + y}

f (x) = - x^{2} + 9, - 3 \leq x \leq 4

f (x) = x^{4} + y^{4} - 4 x y + 2

What is the extreme f(x)=(x^2-9x+50)/(x-7) ?
The extreme f(x)=(x^2-9x+50)/(x-7) is Maximum(1,-7),Minimum(13,17)