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

The extreme f(x)=x^3-3x+1,-3<= x<= 3 is Minimum(-3,-17),Maximum(-1,3),Minimum(1,-1),Maximum(3,19)

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

extreme points

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

Minimum (- 3, - 17), Maximum (- 1, 3), Minimum (1, - 1), Maximum (3, 19)

Solution steps

Find the critical points:

x = - 3, x = - 1, x = 1, x = 3

Domain of

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

Find intervals:

Increasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < 3

Plug the extreme point

x = - 3

into

x^{3} - 3 x + 1 \Rightarrow y = - 17

Minimum (- 3, - 17)

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 3)

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

Plug the extreme point

x = 3

into

x^{3} - 3 x + 1 \Rightarrow y = 19

Maximum (3, 19)

Minimum (- 3, - 17), Maximum (- 1, 3), Minimum (1, - 1), Maximum (3, 19)

23

y = 3 x - 3

- x + 3

f (x) = \frac{- x ^{3}}{5 x ^{2} + 7}

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

What is the extreme f(x)=x^3-3x+1,-3<= x<= 3 ?
The extreme f(x)=x^3-3x+1,-3<= x<= 3 is Minimum(-3,-17),Maximum(-1,3),Minimum(1,-1),Maximum(3,19)