What is the extreme f(x)=2x^4-8x^3+4 ?

The extreme f(x)=2x^4-8x^3+4 is Saddle(0,4),Minimum(3,-50)

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extreme f(x)=2x^4-8x^3+4

extreme points

f (x) = 2 x^{4} - 8 x^{3} + 4

Saddle (0, 4), Minimum (3, - 50)

Solution steps

f^{'} (x) = 8 x^{3} - 24 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 4)

Plug

x = 3

into

2 x^{4} - 8 x^{3} + 4 : - 50

Minimum (3, - 50)

Saddle (0, 4), Minimum (3, - 50)

f (x, y) = 448 x + 233 y + x y - 2 x^{2} - 3 y^{2} - 15438

f (x) = \frac{2 x}{x ^{2} + 3}, - 4 \leq x \leq 0

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

x^{4} - 5 x^{3} + x^{2} + 21 x - 18

x^{2} + 64

What is the extreme f(x)=2x^4-8x^3+4 ?
The extreme f(x)=2x^4-8x^3+4 is Saddle(0,4),Minimum(3,-50)