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

The extreme f(x)=(x^2)/(x-5) is Maximum(0,0),Minimum(10,20)

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

extreme points

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

Maximum (0, 0), Minimum (10, 20)

Solution steps

f^{'} (x) = \frac{x ^{2} - 10 x}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 5,

Decreasing

: 5 < x < 10,

Increasing

: 10 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x - 5} : 0

Maximum (0, 0)

Plug

x = 10

into

\frac{x ^{2}}{x - 5} : 20

Minimum (10, 20)

Maximum (0, 0), Minimum (10, 20)

f (x) = 8 x + 12

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

y = \frac{1}{3} x + 3

g (x) = - x + 3

f (x) = (x - 1)

What is the extreme f(x)=(x^2)/(x-5) ?
The extreme f(x)=(x^2)/(x-5) is Maximum(0,0),Minimum(10,20)