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

The extreme f(x)=2x^3-3x+1 is Maximum(-1/(sqrt(2)),sqrt(2)+1),Minimum(1/(sqrt(2)),-sqrt(2)+1)

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

extreme points

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

Maximum (- \frac{1}{2}, 2 + 1), Minimum (\frac{1}{2}, - 2 + 1)

Solution steps

f^{'} (x) = 6 x^{2} - 3

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

2 x^{3} - 3 x + 1 : 2 + 1

Maximum (- \frac{1}{2}, 2 + 1)

Plug

x = \frac{1}{2}

into

2 x^{3} - 3 x + 1 : - 2 + 1

Minimum (\frac{1}{2}, - 2 + 1)

Maximum (- \frac{1}{2}, 2 + 1), Minimum (\frac{1}{2}, - 2 + 1)

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

f (x) = \frac{8}{x ^{2} + 1}

f (2.52, 5.15) = 3 x^{4} y^{2} - 18 x y

f (x) = - 3 x^{4} + 20 x^{3} - 36 x^{2} + 7

y = - 2 + 5 sin (\frac{π}{12} (x - 2))

What is the extreme f(x)=2x^3-3x+1 ?
The extreme f(x)=2x^3-3x+1 is Maximum(-1/(sqrt(2)),sqrt(2)+1),Minimum(1/(sqrt(2)),-sqrt(2)+1)