What is the extreme f(x)=x^3+3x^2+5,-3<= x<= 2 ?

The extreme f(x)=x^3+3x^2+5,-3<= x<= 2 is Minimum(-3,5),Maximum(-2,9),Minimum(0,5),Maximum(2,25)

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extreme f(x)=x^3+3x^2+5,-3<= x<= 2

extreme points

f (x) = x^{3} + 3 x^{2} + 5, - 3 \leq x \leq 2

Minimum (- 3, 5), Maximum (- 2, 9), Minimum (0, 5), Maximum (2, 25)

Solution steps

Find the critical points:

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

Domain of

x^{3} + 3 x^{2} + 5, - 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} + 5 \Rightarrow y = 5

Minimum (- 3, 5)

Plug the extreme point

x = - 2

into

x^{3} + 3 x^{2} + 5 \Rightarrow y = 9

Maximum (- 2, 9)

Plug the extreme point

x = 0

into

x^{3} + 3 x^{2} + 5 \Rightarrow y = 5

Minimum (0, 5)

Plug the extreme point

x = 2

into

x^{3} + 3 x^{2} + 5 \Rightarrow y = 25

Maximum (2, 25)

Minimum (- 3, 5), Maximum (- 2, 9), Minimum (0, 5), Maximum (2, 25)

2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1)

f (x) = x (26 - 40 + 2 x) (\frac{40}{2} - x)

e^{x^{3} - 12 x}, 0 \leq x \leq 3

f (x, y) = 2 x + 3 y x^{2}

x^{3} + 30 x + 128

What is the extreme f(x)=x^3+3x^2+5,-3<= x<= 2 ?
The extreme f(x)=x^3+3x^2+5,-3<= x<= 2 is Minimum(-3,5),Maximum(-2,9),Minimum(0,5),Maximum(2,25)