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

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

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

extreme points

y = \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)

e^{20 (e^{t} - 1) - 25 t}

(40 x - 0. 5^{2}) - (6 x + 10)

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 6 x + 7 y

f (x) 2 x^{3} + x^{2} - 20 x + 1

f (x) = 2 x^{3} + 3 x^{2} - 72 x + 1, (- 4, 7)

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