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

The extreme f(x)=3x^4-4x^3+6 is Saddle(0,6),Minimum(1,5)

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

extreme points

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

Saddle (0, 6), Minimum (1, 5)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

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

Saddle (0, 6)

Plug

x = 1

into

3 x^{4} - 4 x^{3} + 6 : 5

Minimum (1, 5)

Saddle (0, 6), Minimum (1, 5)

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

f (x, y) = 10 - \frac{x ^{2}}{2} - \frac{y ^{2}}{2}

f (x, y) = x^{2} + y^{2} + x + y + x y

f (x, y) = 6 y^{3} - 5 x^{3} y + 3 x^{2}

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

What is the extreme f(x)=3x^4-4x^3+6 ?
The extreme f(x)=3x^4-4x^3+6 is Saddle(0,6),Minimum(1,5)