What is the extreme f(x)=6x^4-8x^3+2 ?

The extreme f(x)=6x^4-8x^3+2 is Saddle(0,2),Minimum(1,0)

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extreme f(x)=6x^4-8x^3+2

extreme points

f (x) = 6 x^{4} - 8 x^{3} + 2

Saddle (0, 2), Minimum (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

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

Saddle (0, 2)

Plug

x = 1

into

6 x^{4} - 8 x^{3} + 2 : 0

Minimum (1, 0)

Saddle (0, 2), Minimum (1, 0)

f (x, y) = x^{3} - y^{2} - 3 x - 2 y - 2 x y - 3

f (x) = 2 x - \frac{32}{x}

f (x) = 2 x - \frac{4}{x}

f (x) = \frac{( 4 x - 18 )}{( x - 9 ) ^{2}}

f (x, y) = 11 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

What is the extreme f(x)=6x^4-8x^3+2 ?
The extreme f(x)=6x^4-8x^3+2 is Saddle(0,2),Minimum(1,0)