What is the extreme f(x)=(x^2)/(x-4) ?

The extreme f(x)=(x^2)/(x-4) is Maximum(0,0),Minimum(8,16)

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extreme f(x)=(x^2)/(x-4)

extreme points

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

Maximum (0, 0), Minimum (8, 16)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x - 4} : 0

Maximum (0, 0)

Plug

x = 8

into

\frac{x ^{2}}{x - 4} : 16

Minimum (8, 16)

Maximum (0, 0), Minimum (8, 16)

f (x, y) = x^{2} + 4 y^{2}

y = x^{3} - 6 x^{2} + 9 x

f (x) = 2 x^{4} - 16 x^{2} + 20 x - 7

f (x) = 4 x^{3} - 4 x

f (x) = x - 6 x - 1

What is the extreme f(x)=(x^2)/(x-4) ?
The extreme f(x)=(x^2)/(x-4) is Maximum(0,0),Minimum(8,16)