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

The extreme f(x)=x^4+4x^3-1 is Minimum(-3,-28),Saddle(0,-1)

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

extreme points

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

Minimum (- 3, - 28), Saddle (0, - 1)

Solution steps

f^{'} (x) = 4 x^{3} + 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

x^{4} + 4 x^{3} - 1 : - 28

Minimum (- 3, - 28)

Plug

x = 0

into

x^{4} + 4 x^{3} - 1 : - 1

Saddle (0, - 1)

Minimum (- 3, - 28), Saddle (0, - 1)

f (x, y) = x^{2} + x y + y^{2} - 31 y + 320

f (x, y) = 5 x + 3 y

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

f (x) = \frac{( x ^{2} - 7 x + 26 )}{x - 5}

f (x) = x^{3} + 8 x^{2} + 16 x + 8, - 7 \leq x \leq 0

What is the extreme f(x)=x^4+4x^3-1 ?
The extreme f(x)=x^4+4x^3-1 is Minimum(-3,-28),Saddle(0,-1)