What is the extreme f(x)=2x^3-6x+1 ?

The extreme f(x)=2x^3-6x+1 is Maximum(-1,5),Minimum(1,-3)

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extreme f(x)=2x^3-6x+1

extreme points

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

Maximum (- 1, 5), Minimum (1, - 3)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

2 x^{3} - 6 x + 1 : 5

Maximum (- 1, 5)

Plug

x = 1

into

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

Minimum (1, - 3)

Maximum (- 1, 5), Minimum (1, - 3)

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

\frac{7 x}{x ^{2} - 9}

(\frac{1}{2} - x^{2} + y^{2}) e^{1 - x^{2} - y^{2}}

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

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

What is the extreme f(x)=2x^3-6x+1 ?
The extreme f(x)=2x^3-6x+1 is Maximum(-1,5),Minimum(1,-3)