What is the extreme f(x)=-x^3-4.5x^2-6x+2 ?

The extreme f(x)=-x^3-4.5x^2-6x+2 is Minimum(-2,4),Maximum(-1,4.5)

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extreme f(x)=-x^3-4.5x^2-6x+2

extreme points

f (x) = - x^{3} - 4.5 x^{2} - 6 x + 2

Minimum (- 2, 4), Maximum (- 1, 4.5)

Solution steps

f^{'} (x) = - 3 x^{2} - 9 x - 6

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 2

into

- x^{3} - 4.5 x^{2} - 6 x + 2 : 4

Minimum (- 2, 4)

Plug

x = - 1

into

- x^{3} - 4.5 x^{2} - 6 x + 2 : 4.5

Maximum (- 1, 4.5)

Minimum (- 2, 4), Maximum (- 1, 4.5)

f (x, y) = x^{4} + y^{4} + 4 x y

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

5 x + 1 + \frac{2}{x - 2}

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

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

What is the extreme f(x)=-x^3-4.5x^2-6x+2 ?
The extreme f(x)=-x^3-4.5x^2-6x+2 is Minimum(-2,4),Maximum(-1,4.5)