What is the extreme ((x^2-8))/(x-3) ?

The extreme ((x^2-8))/(x-3) is Maximum(2,4),Minimum(4,8)

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extreme ((x^2-8))/(x-3)

extreme points

\frac{( x ^{2} - 8 )}{x - 3}

Maximum (2, 4), Minimum (4, 8)

Solution steps

f^{'} (x) = \frac{x ^{2} - 6 x + 8}{( x - 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 3,

Decreasing

: 3 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 8}{x - 3} : 4

Maximum (2, 4)

Plug

x = 4

into

\frac{x ^{2} - 8}{x - 3} : 8

Minimum (4, 8)

Maximum (2, 4), Minimum (4, 8)

f (x y) = x^{2} + 2 x^{2} - x y + 14 y

y = 21 x + 9 x^{2} - x^{3}

sin (x) cos (x)

f (x) = 2 x^{2} - 8 x + y^{2} - 8 y + 6

f (x) = x^{\frac{1}{5}} (x + 24)

What is the extreme ((x^2-8))/(x-3) ?
The extreme ((x^2-8))/(x-3) is Maximum(2,4),Minimum(4,8)