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) = - 2 x^{2} + 6 x + 10

y = 2 x^{2} - 3 x + 4

f (x) = 4 x + 24

ln (7 - 6 x^{2})

\frac{( x ^{2} - 3 x - 4 )}{x + 2}

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)