What is the extreme f(x)=6x^4-32x^3+48x^2-7 ?

The extreme f(x)=6x^4-32x^3+48x^2-7 is Minimum(0,-7),Saddle(2,25)

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extreme f(x)=6x^4-32x^3+48x^2-7

extreme points

f (x) = 6 x^{4} - 32 x^{3} + 48 x^{2} - 7

Minimum (0, - 7), Saddle (2, 25)

Solution steps

f^{'} (x) = 24 x^{3} - 96 x^{2} + 96 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

6 x^{4} - 32 x^{3} + 48 x^{2} - 7 : - 7

Minimum (0, - 7)

Plug

x = 2

into

6 x^{4} - 32 x^{3} + 48 x^{2} - 7 : 25

Saddle (2, 25)

Minimum (0, - 7), Saddle (2, 25)

What is the extreme f(x)=6x^4-32x^3+48x^2-7 ?
The extreme f(x)=6x^4-32x^3+48x^2-7 is Minimum(0,-7),Saddle(2,25)