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

The extreme f(x)=-x^3-3x^2+9x+1 is Minimum(-3,-26),Maximum(1,6)

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

extreme points

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

Minimum (- 3, - 26), Maximum (1, 6)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 3

into

- x^{3} - 3 x^{2} + 9 x + 1 : - 26

Minimum (- 3, - 26)

Plug

x = 1

into

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

Maximum (1, 6)

Minimum (- 3, - 26), Maximum (1, 6)

y = - x^{2} - 3

\frac{3 x ^{2}}{2 x + 2}

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

f (x) = 2 x e^{5 x}

f (x) = \frac{5 x ^{2}}{x ^{2} - 4}

What is the extreme f(x)=-x^3-3x^2+9x+1 ?
The extreme f(x)=-x^3-3x^2+9x+1 is Minimum(-3,-26),Maximum(1,6)