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

The extreme f(x)=x^4-4x^3+16x is Minimum(-1,-11),Saddle(2,16)

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

extreme points

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

Minimum (- 1, - 11), Saddle (2, 16)

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 : - 11

Minimum (- 1, - 11)

Plug

x = 2

into

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

Saddle (2, 16)

Minimum (- 1, - 11), Saddle (2, 16)

What is the extreme f(x)=x^4-4x^3+16x ?
The extreme f(x)=x^4-4x^3+16x is Minimum(-1,-11),Saddle(2,16)