What is the extreme f(x)=(x^3-4)/(x-1) ?

The extreme f(x)=(x^3-4)/(x-1) is Minimum(-0.91082…,2.48877…)

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extreme f(x)=(x^3-4)/(x-1)

extreme points

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

Minimum (- 0.91082 \dots, 2.48877 \dots)

Solution steps

Find the critical points:

x \approx - 0.91082 \dots

Domain of

\frac{x ^{3} - 4}{x - 1} : x < 1 or x > 1

Find intervals:

Decreasing

: - \infty < x < - 0.91082 \dots,

Increasing

: - 0.91082 \dots < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = - 0.91082 \dots

into

\frac{x ^{3} - 4}{x - 1} \Rightarrow y = 2.48877 \dots

Minimum (- 0.91082 \dots, 2.48877 \dots)

f (x) = - \frac{2}{3} x^{3} + 10 x^{2} - 48 x - 1

f (x) = 4 x - 48 x^{\frac{1}{3}}

f (x, y) = 4 x^{2} + 3 y^{2} + 3 x y - 10 x + 6 y + 36

\frac{x}{x - 1}

e^{x^{2}}

What is the extreme f(x)=(x^3-4)/(x-1) ?
The extreme f(x)=(x^3-4)/(x-1) is Minimum(-0.91082…,2.48877…)