What is the extreme f(x)=x^3-3x+8 ?

The extreme f(x)=x^3-3x+8 is Maximum(-1,10),Minimum(1,6)

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extreme f(x)=x^3-3x+8

extreme points

f (x) = x^{3} - 3 x + 8

Maximum (- 1, 10), Minimum (1, 6)

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 + 8 : 10

Maximum (- 1, 10)

Plug

x = 1

into

x^{3} - 3 x + 8 : 6

Minimum (1, 6)

Maximum (- 1, 10), Minimum (1, 6)

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

f (x) = x^{3} - 3 x - 5

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

f (x) = 4 - 3 x

f (x, y) = x y^{2} - 27 x y - 27 y^{2}

What is the extreme f(x)=x^3-3x+8 ?
The extreme f(x)=x^3-3x+8 is Maximum(-1,10),Minimum(1,6)