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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 3)

Plug

x = 1

into

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

Minimum (1, - 1)

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

\frac{x}{x ^{2} - 1}

f (x) = \frac{x ^{2} - 1}{4 x - 4}

f (x) = \frac{1}{4} x^{2} + 2 x + 6

f (x) = ln (9 - t^{2})

y = \frac{- 7}{1 x - 3}

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