What is the extreme f(x)=(x^3-2)/((x-1)^2) ?

The extreme f(x)=(x^3-2)/((x-1)^2) is Maximum(-1,-3/4),Saddle(2,6)

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extreme f(x)=(x^3-2)/((x-1)^2)

extreme points

f (x) = \frac{x ^{3} - 2}{( x - 1 ) ^{2}}

Maximum (- 1, - \frac{3}{4}), Saddle (2, 6)

Solution steps

f^{'} (x) = \frac{- 3 x ^{2} + x ^{3} + 4}{( x - 1 ) ^{3}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

\frac{x ^{3} - 2}{( x - 1 ) ^{2}} : - \frac{3}{4}

Maximum (- 1, - \frac{3}{4})

Plug

x = 2

into

\frac{x ^{3} - 2}{( x - 1 ) ^{2}} : 6

Saddle (2, 6)

Maximum (- 1, - \frac{3}{4}), Saddle (2, 6)

f (x) = \frac{3}{x + 4}

f (x, y) = ((x + 9)^{2} - 1) (x^{2} - 16) + 200 y^{2} + 2

f (x, y) = 5 x y - 7 x^{2} - y^{2} + 3 x - 6 y + 2

f (x, y) = 3 x e^{y} - x^{3} - e^{3 y}

f (x, y) = 14 x y - x^{3} - 7 y^{2}

What is the extreme f(x)=(x^3-2)/((x-1)^2) ?
The extreme f(x)=(x^3-2)/((x-1)^2) is Maximum(-1,-3/4),Saddle(2,6)