What is the extreme f(x)=x^3-12x+20 ?

The extreme f(x)=x^3-12x+20 is Maximum(-2,36),Minimum(2,4)

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extreme f(x)=x^3-12x+20

extreme points

f (x) = x^{3} - 12 x + 20

Maximum (- 2, 36), Minimum (2, 4)

Solution steps

f^{'} (x) = 3 x^{2} - 12

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{3} - 12 x + 20 : 36

Maximum (- 2, 36)

Plug

x = 2

into

x^{3} - 12 x + 20 : 4

Minimum (2, 4)

Maximum (- 2, 36), Minimum (2, 4)

- \frac{3}{4}

m = \frac{3}{7}

\frac{e ^{x}}{x ^{2}}

f (x) = x^{3} - 27 x + 4

y = 24 x + \frac{1}{( 1 + x ) ^{60}}

What is the extreme f(x)=x^3-12x+20 ?
The extreme f(x)=x^3-12x+20 is Maximum(-2,36),Minimum(2,4)