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

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

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

extreme points

\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)

f (x) = 9 x^{3} + 2 1^{2} + 8 x - 3

- 2 x^{2} + 8 x - 6

f (x) = 9 + 2 x^{2} - x^{4}

f (x) = 3 x^{2} - 6 x + 4, - 1 \leq x \leq 3

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

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