What is the extreme f(x)=x^4-12x^3+16x^2+2 ?

The extreme f(x)=x^4-12x^3+16x^2+2 is Minimum(0,2),Maximum(1,7),Minimum(8,-1022)

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extreme f(x)=x^4-12x^3+16x^2+2

extreme points

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

Minimum (0, 2), Maximum (1, 7), Minimum (8, - 1022)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x^{2} + 32 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

x^{4} - 12 x^{3} + 16 x^{2} + 2 : 2

Minimum (0, 2)

Plug

x = 1

into

x^{4} - 12 x^{3} + 16 x^{2} + 2 : 7

Maximum (1, 7)

Plug

x = 8

into

x^{4} - 12 x^{3} + 16 x^{2} + 2 : - 1022

Minimum (8, - 1022)

Minimum (0, 2), Maximum (1, 7), Minimum (8, - 1022)

What is the extreme f(x)=x^4-12x^3+16x^2+2 ?
The extreme f(x)=x^4-12x^3+16x^2+2 is Minimum(0,2),Maximum(1,7),Minimum(8,-1022)