What is the extreme f(x)=1-9x-6x^2-x^3 ?

The extreme f(x)=1-9x-6x^2-x^3 is Minimum(-3,1),Maximum(-1,5)

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extreme f(x)=1-9x-6x^2-x^3

extreme points

f (x) = 1 - 9 x - 6 x^{2} - x^{3}

Minimum (- 3, 1), Maximum (- 1, 5)

Solution steps

f^{'} (x) = - 3 x^{2} - 12 x - 9

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 3

into

1 - 9 x - 6 x^{2} - x^{3} : 1

Minimum (- 3, 1)

Plug

x = - 1

into

1 - 9 x - 6 x^{2} - x^{3} : 5

Maximum (- 1, 5)

Minimum (- 3, 1), Maximum (- 1, 5)

y = \frac{4 x ^{2}}{x ^{2} - 4}

f (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} + 3 x^{2} + 11

f (x) = - \frac{10}{x ^{2} - 3 x + 14}

y = x^{2} + 3

f (x, y) = \frac{1}{3 ^{- x^{2} - y^{2}} + 1}

What is the extreme f(x)=1-9x-6x^2-x^3 ?
The extreme f(x)=1-9x-6x^2-x^3 is Minimum(-3,1),Maximum(-1,5)