What is the extreme f(x)=3x^4-4x^3-8,-1<= x<= 2 ?

The extreme f(x)=3x^4-4x^3-8,-1<= x<= 2 is Maximum(-1,-1),Saddle(0,-8),Minimum(1,-9),Maximum(2,8)

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extreme f(x)=3x^4-4x^3-8,-1<= x<= 2

extreme points

f (x) = 3 x^{4} - 4 x^{3} - 8, - 1 \leq x \leq 2

Maximum (- 1, - 1), Saddle (0, - 8), Minimum (1, - 9), Maximum (2, 8)

Solution steps

Find the critical points:

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

Domain of

3 x^{4} - 4 x^{3} - 8, - 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} - 8 \Rightarrow y = - 1

Maximum (- 1, - 1)

Plug the extreme point

x = 0

into

3 x^{4} - 4 x^{3} - 8 \Rightarrow y = - 8

Saddle (0, - 8)

Plug the extreme point

x = 1

into

3 x^{4} - 4 x^{3} - 8 \Rightarrow y = - 9

Minimum (1, - 9)

Plug the extreme point

x = 2

into

3 x^{4} - 4 x^{3} - 8 \Rightarrow y = 8

Maximum (2, 8)

Maximum (- 1, - 1), Saddle (0, - 8), Minimum (1, - 9), Maximum (2, 8)

4 x^{4} + 4 y^{4} - 2 x y

K (y, z) = 3 yz

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

f (x) = \frac{x}{x}

f (x) = sec (x), - \frac{π}{6} \leq x \leq \frac{π}{3}

What is the extreme f(x)=3x^4-4x^3-8,-1<= x<= 2 ?
The extreme f(x)=3x^4-4x^3-8,-1<= x<= 2 is Maximum(-1,-1),Saddle(0,-8),Minimum(1,-9),Maximum(2,8)