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

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

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

extreme points

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

f (x) = x^{4} ln (x)

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

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

f (x) = e^{x} (24 - x^{2})

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