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

The extreme f(x)=((x-1)^2)/x is Maximum(-1,-4),Minimum(1,0)

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

extreme points

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

Maximum (- 1, - 4), Minimum (1, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{( x - 1 ) ^{2}}{x} : - 4

Maximum (- 1, - 4)

Plug

x = 1

into

\frac{( x - 1 ) ^{2}}{x} : 0

Minimum (1, 0)

Maximum (- 1, - 4), Minimum (1, 0)

f (x, y) = 4 x y + \frac{2}{x} + \frac{1}{y}

f (x) = x^{4} - 8 x^{3} + 6

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

f (x) = - \frac{6 x}{3 x ^{2} + 8}

f (x, y) = 2 x^{3} + y^{3} + 3 x^{2} - 3 y - 12 x - 4

What is the extreme f(x)=((x-1)^2)/x ?
The extreme f(x)=((x-1)^2)/x is Maximum(-1,-4),Minimum(1,0)