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

The extreme f(x)=((x^2-3))/(x-2) is Maximum(1,2),Minimum(3,6)

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

extreme points

f (x) = \frac{( x ^{2} - 3 )}{x - 2}

Maximum (1, 2), Minimum (3, 6)

Solution steps

f^{'} (x) = \frac{x ^{2} - 4 x + 3}{( x - 2 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 2,

Decreasing

: 2 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

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

Maximum (1, 2)

Plug

x = 3

into

\frac{x ^{2} - 3}{x - 2} : 6

Minimum (3, 6)

Maximum (1, 2), Minimum (3, 6)

f (x) = 4 x + \frac{9}{x}

f (x) = x^{3} + 3 x^{2} + 3 x

f (x) = \frac{1}{4} x^{4} - x^{3} - 4 x + 2

120 x - 2 x^{2} - x y - 3 y^{2} + 160 y + 7

f (x) = ∣ x (x - 1) ∣ o n [0, 1]

What is the extreme f(x)=((x^2-3))/(x-2) ?
The extreme f(x)=((x^2-3))/(x-2) is Maximum(1,2),Minimum(3,6)