What is the extreme f(x)=x^3-3x+9 ?

The extreme f(x)=x^3-3x+9 is Maximum(-1,11),Minimum(1,7)

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extreme f(x)=x^3-3x+9

extreme points

f (x) = x^{3} - 3 x + 9

Maximum (- 1, 11), Minimum (1, 7)

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 + 9 : 11

Maximum (- 1, 11)

Plug

x = 1

into

x^{3} - 3 x + 9 : 7

Minimum (1, 7)

Maximum (- 1, 11), Minimum (1, 7)

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

+ \frac{1}{11 - t}

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

- 5 x^{6} ln (x)

f (x) = 4 x^{3} - 6 x^{2} y + 12 x^{2} - 12 y^{2}

What is the extreme f(x)=x^3-3x+9 ?
The extreme f(x)=x^3-3x+9 is Maximum(-1,11),Minimum(1,7)