What is the extreme f(x)=3+(\sqrt[3]{x^3-1})/(x-3) ?

The extreme f(x)=3+(\sqrt[3]{x^3-1})/(x-3) is Minimum(-0.57735…,3.29642…),Maximum(0.57735…,3.38438…)

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extreme f(x)=3+(\sqrt[3]{x^3-1})/(x-3)

extreme points

Minimum (- 0.57735 \dots, 3.29642 \dots), Maximum (0.57735 \dots, 3.38438 \dots)

Solution steps

Find the critical points:

x \approx - 0.57735 \dots, x \approx 0.57735 \dots, x = 1

Domain of

Find intervals:

Decreasing

: - \infty < x < - 0.57735 \dots,

Increasing

: - 0.57735 \dots < x < 0.57735 \dots,

Decreasing

: 0.57735 \dots < x < 1,

Decreasing

: 1 < x < 3,

Decreasing

: 3 < x < \infty

Plug the extreme point

x = - 0.57735 \dots

into

Minimum (- 0.57735 \dots, 3.29642 \dots)

Plug the extreme point

x = 0.57735 \dots

into

Maximum (0.57735 \dots, 3.38438 \dots)

Minimum (- 0.57735 \dots, 3.29642 \dots), Maximum (0.57735 \dots, 3.38438 \dots)

f (x) = - 2 x^{3} + 21 x^{2} - 36 x + 3

f (x) = 3 x^{2} + 5 y^{2} - 10 x y - 6 y + 1

9600 + 280 x - 10 x^{2}

4 x^{3} + 9 x^{2} - 54 x + 6

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

What is the extreme f(x)=3+(\sqrt[3]{x^3-1})/(x-3) ?
The extreme f(x)=3+(\sqrt[3]{x^3-1})/(x-3) is Minimum(-0.57735…,3.29642…),Maximum(0.57735…,3.38438…)