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

The extreme f(x)=3x^4-4x^3,-1<= x<= 2 is Maximum(-1,7),Saddle(0,0),Minimum(1,-1),Maximum(2,16)

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

extreme points

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

Maximum (- 1, 7), Saddle (0, 0), Minimum (1, - 1), Maximum (2, 16)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 2

Domain of

3 x^{4} - 4 x^{3}, - 1 \leq x \leq 2 : - 1 \leq x \leq 2

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 1

into

3 x^{4} - 4 x^{3} \Rightarrow y = 7

Maximum (- 1, 7)

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)

Plug the extreme point

x = 2

into

3 x^{4} - 4 x^{3} \Rightarrow y = 16

Maximum (2, 16)

Maximum (- 1, 7), Saddle (0, 0), Minimum (1, - 1), Maximum (2, 16)

f (x) = x^{3} + 45 x^{2} + 87 x + 28

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

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

f (x) = \frac{- 6}{x ^{2} - 9}

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

What is the extreme f(x)=3x^4-4x^3,-1<= x<= 2 ?
The extreme f(x)=3x^4-4x^3,-1<= x<= 2 is Maximum(-1,7),Saddle(0,0),Minimum(1,-1),Maximum(2,16)