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) = (7 - 3) \cdot e^{3 x}

x^{4} + 2 x^{3} + 2 x + 2

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

f (x, y) = (2 x + 3 y) (5 x + 7 y)

- 2 x^{4} + x^{3} + 6 x^{2} + x - 2

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)