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

The extreme f(x)=2x^4-8x^3 is Saddle(0,0),Minimum(3,-54)

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

extreme points

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

Saddle (0, 0), Minimum (3, - 54)

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} : 0

Saddle (0, 0)

Plug

x = 3

into

2 x^{4} - 8 x^{3} : - 54

Minimum (3, - 54)

Saddle (0, 0), Minimum (3, - 54)

f (x) = - 5 x^{2} + 40 x

y = x^{2} - x - 42

f (x) = - \frac{1}{3} x

2 x - 3 y = 9, (2, - 1)

2 + \frac{1}{x}

What is the extreme f(x)=2x^4-8x^3 ?
The extreme f(x)=2x^4-8x^3 is Saddle(0,0),Minimum(3,-54)