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

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

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

extreme points

y = \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)

6 + 2 x - 2 x^{2}

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

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

x^{2} - 2 x - 15

f (x) = x ln (\frac{x}{3})

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