What is the extreme f(x)=(x^2-9)/(x-5) ?

The extreme f(x)=(x^2-9)/(x-5) is Maximum(1,2),Minimum(9,18)

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extreme f(x)=(x^2-9)/(x-5)

extreme points

f (x) = \frac{x ^{2} - 9}{x - 5}

Maximum (1, 2), Minimum (9, 18)

Solution steps

f^{'} (x) = \frac{x ^{2} - 10 x + 9}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 5,

Decreasing

: 5 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 9}{x - 5} : 2

Maximum (1, 2)

Plug

x = 9

into

\frac{x ^{2} - 9}{x - 5} : 18

Minimum (9, 18)

Maximum (1, 2), Minimum (9, 18)

y + 3 = \frac{1}{2} (x + 10)

lo g_{\frac{1}{3}} (x)

f (3) = \frac{9}{3 - 10 x} - 3

f (x) = 9 cos (\frac{1}{4} π x - π) - 2

8 x + 10 y = - 60

What is the extreme f(x)=(x^2-9)/(x-5) ?
The extreme f(x)=(x^2-9)/(x-5) is Maximum(1,2),Minimum(9,18)