What is the extreme f(x)=3x^4-4x^3,(-1,2) ?

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

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extreme f(x)=3x^4-4x^3,(-1,2)

extreme points

f (x) = 3 x^{4} - 4 x^{3}, (- 1, 2)

Saddle (0, 0), Minimum (1, - 1)

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

3 x^{4} - 4 x^{3}, (- 1, 2) : - 1 < x < 2

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 0

into

3 x^{4} - 4 x^{3} \Rightarrow y = 0

Saddle (0, 0)

Plug the extreme point

x = 1

into

3 x^{4} - 4 x^{3} \Rightarrow y = - 1

Minimum (1, - 1)

Saddle (0, 0), Minimum (1, - 1)

2 x ln (x) + x

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

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

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

f (x) = 4 x^{3} - 39 x^{2} + 90 x + 2, 1 \leq x \leq 6

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