What is the extreme f(x)=x^4-8x^3+8 ?

The extreme f(x)=x^4-8x^3+8 is Saddle(0,8),Minimum(6,-424)

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extreme f(x)=x^4-8x^3+8

extreme points

f (x) = x^{4} - 8 x^{3} + 8

Saddle (0, 8), Minimum (6, - 424)

Solution steps

f^{'} (x) = 4 x^{3} - 24 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{4} - 8 x^{3} + 8 : 8

Saddle (0, 8)

Plug

x = 6

into

x^{4} - 8 x^{3} + 8 : - 424

Minimum (6, - 424)

Saddle (0, 8), Minimum (6, - 424)

2 x

f (x) = \frac{1}{6} (12 - 7 x) + 1

f (x) = 4 x^{2} + 2

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

f (x) = \frac{1}{5 x}

What is the extreme f(x)=x^4-8x^3+8 ?
The extreme f(x)=x^4-8x^3+8 is Saddle(0,8),Minimum(6,-424)