What is the extreme f(x)=2x^3-9x^2+12x+6,0<= x<= 3 ?

The extreme f(x)=2x^3-9x^2+12x+6,0<= x<= 3 is Minimum(0,6),Maximum(1,11),Minimum(2,10),Maximum(3,15)

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extreme f(x)=2x^3-9x^2+12x+6,0<= x<= 3

extreme points

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6, 0 \leq x \leq 3

Minimum (0, 6), Maximum (1, 11), Minimum (2, 10), Maximum (3, 15)

Solution steps

Find the critical points:

x = 0, x = 1, x = 2, x = 3

Domain of

2 x^{3} - 9 x^{2} + 12 x + 6, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

2 x^{3} - 9 x^{2} + 12 x + 6 \Rightarrow y = 6

Minimum (0, 6)

Plug the extreme point

x = 1

into

2 x^{3} - 9 x^{2} + 12 x + 6 \Rightarrow y = 11

Maximum (1, 11)

Plug the extreme point

x = 2

into

2 x^{3} - 9 x^{2} + 12 x + 6 \Rightarrow y = 10

Minimum (2, 10)

Plug the extreme point

x = 3

into

2 x^{3} - 9 x^{2} + 12 x + 6 \Rightarrow y = 15

Maximum (3, 15)

Minimum (0, 6), Maximum (1, 11), Minimum (2, 10), Maximum (3, 15)

y = \frac{x}{x ^{2} - x + 4}

\frac{4 x}{1 + 9 x ^{2}}

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

f (x) = \frac{1}{x}, (- 2, 3)

y = x - ln (x)

What is the extreme f(x)=2x^3-9x^2+12x+6,0<= x<= 3 ?
The extreme f(x)=2x^3-9x^2+12x+6,0<= x<= 3 is Minimum(0,6),Maximum(1,11),Minimum(2,10),Maximum(3,15)