What is the extreme f(x)=2x^3+3x^2-12x+8,-2<= x<= 4 ?

The extreme f(x)=2x^3+3x^2-12x+8,-2<= x<= 4 is Maximum(-2,28),Minimum(1,1),Maximum(4,136)

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extreme f(x)=2x^3+3x^2-12x+8,-2<= x<= 4

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8, - 2 \leq x \leq 4

Maximum (- 2, 28), Minimum (1, 1), Maximum (4, 136)

Solution steps

Find the critical points:

x = - 2, x = 1, x = 4

Domain of

2 x^{3} + 3 x^{2} - 12 x + 8, - 2 \leq x \leq 4 : - 2 \leq x \leq 4

Find intervals:

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < 4

Plug the extreme point

x = - 2

into

2 x^{3} + 3 x^{2} - 12 x + 8 \Rightarrow y = 28

Maximum (- 2, 28)

Plug the extreme point

x = 1

into

2 x^{3} + 3 x^{2} - 12 x + 8 \Rightarrow y = 1

Minimum (1, 1)

Plug the extreme point

x = 4

into

2 x^{3} + 3 x^{2} - 12 x + 8 \Rightarrow y = 136

Maximum (4, 136)

Maximum (- 2, 28), Minimum (1, 1), Maximum (4, 136)

f (x) = 2 x^{3} - 3 x^{2} - 12 + 1, - 2 \leq x \leq 3

2 x^{2} - 2 (x - 5)

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

- 2 x^{2} + 4 x

f (x, y) = (x - 0.5)^{2} - y^{2}

What is the extreme f(x)=2x^3+3x^2-12x+8,-2<= x<= 4 ?
The extreme f(x)=2x^3+3x^2-12x+8,-2<= x<= 4 is Maximum(-2,28),Minimum(1,1),Maximum(4,136)