What is the extreme f(x)=x(2x^3+3x-12) ?

The extreme f(x)=x(2x^3+3x-12) is Minimum(0.92944…,-7.06920…)

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extreme f(x)=x(2x^3+3x-12)

extreme points

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

Minimum (0.92944 \dots, - 7.06920 \dots)

Solution steps

Find the critical points:

x \approx 0.92944 \dots

Domain of

x (2 x^{3} + 3 x - 12) : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0.92944 \dots,

Increasing

: 0.92944 \dots < x < \infty

Plug the extreme point

x = 0.92944 \dots

into

x (2 x^{3} + 3 x - 12) \Rightarrow y = - 7.06920 \dots

Minimum (0.92944 \dots, - 7.06920 \dots)

f (x) = 10 + 3 x - x^{2}

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

f (x) = \frac{x ^{2} - 11 x + 82}{x - 9}

x^{3} - 12 x^{2} + 45 x + 1

10

What is the extreme f(x)=x(2x^3+3x-12) ?
The extreme f(x)=x(2x^3+3x-12) is Minimum(0.92944…,-7.06920…)