What is the extreme f(x)=x^3+3x^2+9,-3<= x<= 2 ?

The extreme f(x)=x^3+3x^2+9,-3<= x<= 2 is Minimum(-3,9),Maximum(-2,13),Minimum(0,9),Maximum(2,29)

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extreme f(x)=x^3+3x^2+9,-3<= x<= 2

extreme points

f (x) = x^{3} + 3 x^{2} + 9, - 3 \leq x \leq 2

Minimum (- 3, 9), Maximum (- 2, 13), Minimum (0, 9), Maximum (2, 29)

Solution steps

Find the critical points:

x = - 3, x = - 2, x = 0, x = 2

Domain of

x^{3} + 3 x^{2} + 9, - 3 \leq x \leq 2 : - 3 \leq x \leq 2

Find intervals:

Increasing

: - 3 < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 3

into

x^{3} + 3 x^{2} + 9 \Rightarrow y = 9

Minimum (- 3, 9)

Plug the extreme point

x = - 2

into

x^{3} + 3 x^{2} + 9 \Rightarrow y = 13

Maximum (- 2, 13)

Plug the extreme point

x = 0

into

x^{3} + 3 x^{2} + 9 \Rightarrow y = 9

Minimum (0, 9)

Plug the extreme point

x = 2

into

x^{3} + 3 x^{2} + 9 \Rightarrow y = 29

Maximum (2, 29)

Minimum (- 3, 9), Maximum (- 2, 13), Minimum (0, 9), Maximum (2, 29)

f (x) = \frac{( x ^{2} + x + 1 )}{x ^{2}}

f (x) = x^{3} \cdot e^{x - 1}

y = x^{3} + x^{2} - x

f (x) = \frac{6 x ^{2}}{x - 6}, - 3 \leq x \leq 2

f (x) = x - 3 x

What is the extreme f(x)=x^3+3x^2+9,-3<= x<= 2 ?
The extreme f(x)=x^3+3x^2+9,-3<= x<= 2 is Minimum(-3,9),Maximum(-2,13),Minimum(0,9),Maximum(2,29)