What is the extreme f(x)=4x^3-3x^2-6x+3,0<= x<= 10 ?

The extreme f(x)=4x^3-3x^2-6x+3,0<= x<= 10 is Maximum(0,3),Minimum(1,-2),Maximum(10,3643)

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extreme f(x)=4x^3-3x^2-6x+3,0<= x<= 10

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, 0 \leq x \leq 10

Maximum (0, 3), Minimum (1, - 2), Maximum (10, 3643)

Solution steps

Find the critical points:

x = 0, x = 1, x = 10

Domain of

4 x^{3} - 3 x^{2} - 6 x + 3, 0 \leq x \leq 10 : 0 \leq x \leq 10

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 10

Plug the extreme point

x = 0

into

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

Maximum (0, 3)

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

Plug the extreme point

x = 10

into

4 x^{3} - 3 x^{2} - 6 x + 3 \Rightarrow y = 3643

Maximum (10, 3643)

Maximum (0, 3), Minimum (1, - 2), Maximum (10, 3643)

What is the extreme f(x)=4x^3-3x^2-6x+3,0<= x<= 10 ?
The extreme f(x)=4x^3-3x^2-6x+3,0<= x<= 10 is Maximum(0,3),Minimum(1,-2),Maximum(10,3643)