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

The extreme f(x)=3x^4-8x^3+8 is Saddle(0,8),Minimum(2,-8)

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

extreme points

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

Saddle (0, 8), Minimum (2, - 8)

Solution steps

f^{'} (x) = 12 x^{3} - 24 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Saddle (0, 8)

Plug

x = 2

into

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

Minimum (2, - 8)

Saddle (0, 8), Minimum (2, - 8)

f (x) = x^{2} - 6 x + 4

f (x) = \frac{x + 4}{5 - 3 x}

f (x) = \frac{3 x - 5}{7}

x + 3

f (x) = (4 - x) \cdot e^{2 x}

What is the extreme f(x)=3x^4-8x^3+8 ?
The extreme f(x)=3x^4-8x^3+8 is Saddle(0,8),Minimum(2,-8)