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

The extreme f(x)=x^4-4x^3+8 is Saddle(0,8),Minimum(3,-19)

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

extreme points

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

Saddle (0, 8), Minimum (3, - 19)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 8)

Plug

x = 3

into

x^{4} - 4 x^{3} + 8 : - 19

Minimum (3, - 19)

Saddle (0, 8), Minimum (3, - 19)

y = x

g (x) = \frac{x}{5 + x ^{2}}

f (x) = 5 - 8 x + 2 x^{2}

3 tan^{2} (x)

f (x) = \frac{3}{2} x + 12

What is the extreme f(x)=x^4-4x^3+8 ?
The extreme f(x)=x^4-4x^3+8 is Saddle(0,8),Minimum(3,-19)