What is the extreme f(x)=-3x^4-8x^3-6x^2+1 ?

The extreme f(x)=-3x^4-8x^3-6x^2+1 is Saddle(-1,0),Maximum(0,1)

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extreme f(x)=-3x^4-8x^3-6x^2+1

extreme points

f (x) = - 3 x^{4} - 8 x^{3} - 6 x^{2} + 1

Saddle (- 1, 0), Maximum (0, 1)

Solution steps

f^{'} (x) = - 12 x^{3} - 24 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 1

into

- 3 x^{4} - 8 x^{3} - 6 x^{2} + 1 : 0

Saddle (- 1, 0)

Plug

x = 0

into

- 3 x^{4} - 8 x^{3} - 6 x^{2} + 1 : 1

Maximum (0, 1)

Saddle (- 1, 0), Maximum (0, 1)

f (x) = 3 x^{2} + 16 x

f (x) = 8 + 5 x

cos (2 φ)

y + 3 = \frac{3}{4} x + 5

f (x) = ln (x) + 6

What is the extreme f(x)=-3x^4-8x^3-6x^2+1 ?
The extreme f(x)=-3x^4-8x^3-6x^2+1 is Saddle(-1,0),Maximum(0,1)