What is the extreme f(x)=3x^4-4x^3+5 ?

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

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extreme f(x)=3x^4-4x^3+5

extreme points

f (x) = 3 x^{4} - 4 x^{3} + 5

Saddle (0, 5), Minimum (1, 4)

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} + 5 : 5

Saddle (0, 5)

Plug

x = 1

into

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

Minimum (1, 4)

Saddle (0, 5), Minimum (1, 4)

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

f (x, y) = x y e^{- 8 y}

f (x, y) = x^{2} - x y + y^{2} + 4 x + 7 y

x^{3} + y^{3} - 6 y^{2} - 3 x + 9

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