What is the extreme f(x)=2x^3-3x^2-12x+5,0<= x<= 3 ?

The extreme f(x)=2x^3-3x^2-12x+5,0<= x<= 3 is Maximum(0,5),Minimum(2,-15),Maximum(3,-4)

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extreme f(x)=2x^3-3x^2-12x+5,0<= x<= 3

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 5, 0 \leq x \leq 3

Maximum (0, 5), Minimum (2, - 15), Maximum (3, - 4)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

2 x^{3} - 3 x^{2} - 12 x + 5, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

2 x^{3} - 3 x^{2} - 12 x + 5 \Rightarrow y = 5

Maximum (0, 5)

Plug the extreme point

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 5 \Rightarrow y = - 15

Minimum (2, - 15)

Plug the extreme point

x = 3

into

2 x^{3} - 3 x^{2} - 12 x + 5 \Rightarrow y = - 4

Maximum (3, - 4)

Maximum (0, 5), Minimum (2, - 15), Maximum (3, - 4)

- x^{2} + 3 x - 2, 1 \leq x \leq 3

\frac{2 x - x ^{2}}{x}

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

f (x) = \frac{x ^{2} - 27}{x - 6}

f (x) = - 2 x^{3} + 3 x^{2} + 12 x + 2

What is the extreme f(x)=2x^3-3x^2-12x+5,0<= x<= 3 ?
The extreme f(x)=2x^3-3x^2-12x+5,0<= x<= 3 is Maximum(0,5),Minimum(2,-15),Maximum(3,-4)