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

The extreme f(x)=x^3-3x+1,0<= x<= 5 is Maximum(0,1),Minimum(1,-1),Maximum(5,111)

extreme f(x)=x^3-3x+1,0<= x<= 5

extreme points

f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 5

Maximum (0, 1), Minimum (1, - 1), Maximum (5, 111)

Solution steps

Find the critical points:

x = 0, x = 1, x = 5

Domain of

x^{3} - 3 x + 1, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 5

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

Plug the extreme point

x = 5

into

x^{3} - 3 x + 1 \Rightarrow y = 111

Maximum (5, 111)

Maximum (0, 1), Minimum (1, - 1), Maximum (5, 111)

f (x) = cos (7 x)

f (x) = x - 2 ln (x)

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

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

f (x) = 1 - x

What is the extreme f(x)=x^3-3x+1,0<= x<= 5 ?
The extreme f(x)=x^3-3x+1,0<= x<= 5 is Maximum(0,1),Minimum(1,-1),Maximum(5,111)