What is the extreme f(x)=2x^3+3x^2-12x+4,-2<= x<= 5 ?

The extreme f(x)=2x^3+3x^2-12x+4,-2<= x<= 5 is Maximum(-2,24),Minimum(1,-3),Maximum(5,269)

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extreme f(x)=2x^3+3x^2-12x+4,-2<= x<= 5

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4, - 2 \leq x \leq 5

Maximum (- 2, 24), Minimum (1, - 3), Maximum (5, 269)

Solution steps

Find the critical points:

x = - 2, x = 1, x = 5

Domain of

2 x^{3} + 3 x^{2} - 12 x + 4, - 2 \leq x \leq 5 : - 2 \leq x \leq 5

Find intervals:

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < 5

Plug the extreme point

x = - 2

into

2 x^{3} + 3 x^{2} - 12 x + 4 \Rightarrow y = 24

Maximum (- 2, 24)

Plug the extreme point

x = 1

into

2 x^{3} + 3 x^{2} - 12 x + 4 \Rightarrow y = - 3

Minimum (1, - 3)

Plug the extreme point

x = 5

into

2 x^{3} + 3 x^{2} - 12 x + 4 \Rightarrow y = 269

Maximum (5, 269)

Maximum (- 2, 24), Minimum (1, - 3), Maximum (5, 269)

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

f (x) = x^{2} - 12 x + 5

f (x) = x^{2} - y^{2} + 21

f (x, y) = (9 x^{2} + 4 x + 5) (9 y^{2} + 6 y + 6)

f (x) = 6 x^{3} - x^{2} - 5 x + 7

What is the extreme f(x)=2x^3+3x^2-12x+4,-2<= x<= 5 ?
The extreme f(x)=2x^3+3x^2-12x+4,-2<= x<= 5 is Maximum(-2,24),Minimum(1,-3),Maximum(5,269)