What is the extreme f(x)=(x^2-5)/(x-3) ?

The extreme f(x)=(x^2-5)/(x-3) is Maximum(1,2),Minimum(5,10)

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extreme f(x)=(x^2-5)/(x-3)

extreme points

f (x) = \frac{x ^{2} - 5}{x - 3}

Maximum (1, 2), Minimum (5, 10)

Solution steps

f^{'} (x) = \frac{x ^{2} - 6 x + 5}{( x - 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Decreasing

: 3 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 5}{x - 3} : 2

Maximum (1, 2)

Plug

x = 5

into

\frac{x ^{2} - 5}{x - 3} : 10

Minimum (5, 10)

Maximum (1, 2), Minimum (5, 10)

f (x) = x^{2} - 2 x, - 1 \leq x \leq 2

f (x) = - x^{3} + 3 x^{2} - 6, - 2 \leq x \leq 5

\frac{2 x - 2}{x ^{2} + x} + 1

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

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

What is the extreme f(x)=(x^2-5)/(x-3) ?
The extreme f(x)=(x^2-5)/(x-3) is Maximum(1,2),Minimum(5,10)