What is the extreme f(x)=((x^2-63))/(x-8) ?

The extreme f(x)=((x^2-63))/(x-8) is Maximum(7,14),Minimum(9,18)

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extreme f(x)=((x^2-63))/(x-8)

extreme points

f (x) = \frac{( x ^{2} - 63 )}{x - 8}

Maximum (7, 14), Minimum (9, 18)

Solution steps

f^{'} (x) = \frac{x ^{2} - 16 x + 63}{( x - 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 7,

Decreasing

: 7 < x < 8,

Decreasing

: 8 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 7

into

\frac{x ^{2} - 63}{x - 8} : 14

Maximum (7, 14)

Plug

x = 9

into

\frac{x ^{2} - 63}{x - 8} : 18

Minimum (9, 18)

Maximum (7, 14), Minimum (9, 18)

What is the extreme f(x)=((x^2-63))/(x-8) ?
The extreme f(x)=((x^2-63))/(x-8) is Maximum(7,14),Minimum(9,18)