What is the extreme f(x)=4x^3-12x^2-96x+9 ?

The extreme f(x)=4x^3-12x^2-96x+9 is Maximum(-2,121),Minimum(4,-311)

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extreme f(x)=4x^3-12x^2-96x+9

extreme points

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 9

Maximum (- 2, 121), Minimum (4, - 311)

Solution steps

f^{'} (x) = 12 x^{2} - 24 x - 96

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

4 x^{3} - 12 x^{2} - 96 x + 9 : 121

Maximum (- 2, 121)

Plug

x = 4

into

4 x^{3} - 12 x^{2} - 96 x + 9 : - 311

Minimum (4, - 311)

Maximum (- 2, 121), Minimum (4, - 311)

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 7

f (x, y) = - 25 x^{2} + 250 x - y^{2} + 8 y - 637

f (x, y) = 5 x^{2} y - 5 x y^{3} + 4 y^{3}

f (x) = 12 x - 6

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

What is the extreme f(x)=4x^3-12x^2-96x+9 ?
The extreme f(x)=4x^3-12x^2-96x+9 is Maximum(-2,121),Minimum(4,-311)