What is the extreme f(x)=(x^2)/(sqrt(x)-1) ?

The extreme f(x)=(x^2)/(sqrt(x)-1) is Maximum(0,0),Minimum(16/9 , 256/27)

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extreme f(x)=(x^2)/(sqrt(x)-1)

extreme points

f (x) = \frac{x ^{2}}{x - 1}

Maximum (0, 0), Minimum (\frac{16}{9}, \frac{256}{27})

Solution steps

Find the critical points:

x = 0, x = \frac{16}{9}

Domain of

\frac{x ^{2}}{x - 1} : 0 \leq x < 1 or x > 1

Find intervals:

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < \frac{16}{9},

Increasing

: \frac{16}{9} < x < \infty

Plug the extreme point

x = 0

into

\frac{x ^{2}}{x - 1} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = \frac{16}{9}

into

\frac{x ^{2}}{x - 1} \Rightarrow y = \frac{256}{27}

Minimum (\frac{16}{9}, \frac{256}{27})

Maximum (0, 0), Minimum (\frac{16}{9}, \frac{256}{27})

P (- 8.5) Q (- 4.11)

x^{3} + 8 x^{2} - 6 x - 4

f (x) = - \frac{13}{9} x^{2} + \frac{8}{3} x^{2} + \frac{37}{10}

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

f (x) = \frac{2 x - 6}{x ^{2} + 7 - 4}

What is the extreme f(x)=(x^2)/(sqrt(x)-1) ?
The extreme f(x)=(x^2)/(sqrt(x)-1) is Maximum(0,0),Minimum(16/9 , 256/27)