What is the extreme 2(x-1)(x^2+1)(3x^2-2x+1) ?

The extreme 2(x-1)(x^2+1)(3x^2-2x+1) is Maximum(0.46612…,-0.93525…),Minimum(0.65382…,-0.96342…)

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extreme 2(x-1)(x^2+1)(3x^2-2x+1)

extreme points

2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1)

Maximum (0.46612 \dots, - 0.93525 \dots), Minimum (0.65382 \dots, - 0.96342 \dots)

Solution steps

Find the critical points:

x \approx 0.46612 \dots, x \approx 0.65382 \dots

Domain of

2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1) : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < 0.46612 \dots,

Decreasing

: 0.46612 \dots < x < 0.65382 \dots,

Increasing

: 0.65382 \dots < x < \infty

Plug the extreme point

x = 0.46612 \dots

into

2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1) \Rightarrow y = - 0.93525 \dots

Maximum (0.46612 \dots, - 0.93525 \dots)

Plug the extreme point

x = 0.65382 \dots

into

2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1) \Rightarrow y = - 0.96342 \dots

Minimum (0.65382 \dots, - 0.96342 \dots)

Maximum (0.46612 \dots, - 0.93525 \dots), Minimum (0.65382 \dots, - 0.96342 \dots)

f (x) = x (26 - 40 + 2 x) (\frac{40}{2} - x)

e^{x^{3} - 12 x}, 0 \leq x \leq 3

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

x^{3} + 30 x + 128

f (x) = (x - 5) e^{x}

What is the extreme 2(x-1)(x^2+1)(3x^2-2x+1) ?
The extreme 2(x-1)(x^2+1)(3x^2-2x+1) is Maximum(0.46612…,-0.93525…),Minimum(0.65382…,-0.96342…)