What is the extreme f(x)=x^3-6x+1 ?

The extreme f(x)=x^3-6x+1 is Maximum(-sqrt(2),4sqrt(2)+1),Minimum(sqrt(2),-4sqrt(2)+1)

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extreme f(x)=x^3-6x+1

extreme points

f (x) = x^{3} - 6 x + 1

Maximum (- 2, 42 + 1), Minimum (2, - 42 + 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{3} - 6 x + 1 : 42 + 1

Maximum (- 2, 42 + 1)

Plug

x = 2

into

x^{3} - 6 x + 1 : - 42 + 1

Minimum (2, - 42 + 1)

Maximum (- 2, 42 + 1), Minimum (2, - 42 + 1)

3 x^{3} + y^{2} - 9 x + 4 y

f (x) = 5 x^{6} - 6 x^{5} + 1

f (x) = 3 x - x^{3} - 2 y^{2} + y^{4}

f (x) = 4 x^{4} - 2 (4)^{2} x^{3}

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

What is the extreme f(x)=x^3-6x+1 ?
The extreme f(x)=x^3-6x+1 is Maximum(-sqrt(2),4sqrt(2)+1),Minimum(sqrt(2),-4sqrt(2)+1)