What is the extreme f(x)=x^4-4x^3+16x+4 ?

The extreme f(x)=x^4-4x^3+16x+4 is Minimum(-1,-7),Saddle(2,20)

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extreme f(x)=x^4-4x^3+16x+4

extreme points

f (x) = x^{4} - 4 x^{3} + 16 x + 4

Minimum (- 1, - 7), Saddle (2, 20)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2} + 16

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

x^{4} - 4 x^{3} + 16 x + 4 : - 7

Minimum (- 1, - 7)

Plug

x = 2

into

x^{4} - 4 x^{3} + 16 x + 4 : 20

Saddle (2, 20)

Minimum (- 1, - 7), Saddle (2, 20)

f (x) = x^{7} - 7 x

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

y = 2 x^{2} - 8 x + 8

f (x, y) = (4 x^{2} + 10 y^{2}) e^{- y}

P (x, y) = 6 x^{2} - x y - y^{2} + y + 8 x + 2

What is the extreme f(x)=x^4-4x^3+16x+4 ?
The extreme f(x)=x^4-4x^3+16x+4 is Minimum(-1,-7),Saddle(2,20)