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

The extreme f(x)=x^3-3x+3,0<= x<= 2 is Maximum(0,3),Minimum(1,1),Maximum(2,5)

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

extreme points

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

Maximum (0, 3), Minimum (1, 1), Maximum (2, 5)

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

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

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{3} - 3 x + 3 \Rightarrow y = 3

Maximum (0, 3)

Plug the extreme point

x = 1

into

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

Minimum (1, 1)

Plug the extreme point

x = 2

into

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

Maximum (2, 5)

Maximum (0, 3), Minimum (1, 1), Maximum (2, 5)

f (x, y) = 2 x y - \frac{1}{2} (x^{4} + y^{4}) + 1

f (x, y) = x + y + \frac{1}{x} + \frac{4}{y}

3 x^{4} - 7 x^{2} + 6 x^{2} - 5 x - 8

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

P (x, y) = 200 x^{\frac{1}{2}} y^{\frac{1}{2}}

What is the extreme f(x)=x^3-3x+3,0<= x<= 2 ?
The extreme f(x)=x^3-3x+3,0<= x<= 2 is Maximum(0,3),Minimum(1,1),Maximum(2,5)