What is the extreme f(x)=(x^2)/(x-8) ?

The extreme f(x)=(x^2)/(x-8) is Maximum(0,0),Minimum(16,32)

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extreme f(x)=(x^2)/(x-8)

extreme points

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

Maximum (0, 0), Minimum (16, 32)

Solution steps

f^{'} (x) = \frac{x ^{2} - 16 x}{( x - 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Decreasing

: 8 < x < 16,

Increasing

: 16 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x - 8} : 0

Maximum (0, 0)

Plug

x = 16

into

\frac{x ^{2}}{x - 8} : 32

Minimum (16, 32)

Maximum (0, 0), Minimum (16, 32)

3 (x - 9)

f (x) = \frac{t - 2}{t + 2}

f (x) = - \frac{5}{2 x ^{\frac{3}{2}}}

x (4 - x)

lo g_{2} (x - 3)

What is the extreme f(x)=(x^2)/(x-8) ?
The extreme f(x)=(x^2)/(x-8) is Maximum(0,0),Minimum(16,32)