What is the extreme f(x)=x^3+3x^2+4,-3<= x<= 2 ?

The extreme f(x)=x^3+3x^2+4,-3<= x<= 2 is Minimum(-3,4),Maximum(-2,8),Minimum(0,4),Maximum(2,24)

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extreme f(x)=x^3+3x^2+4,-3<= x<= 2

extreme points

f (x) = x^{3} + 3 x^{2} + 4, - 3 \leq x \leq 2

Minimum (- 3, 4), Maximum (- 2, 8), Minimum (0, 4), Maximum (2, 24)

Solution steps

Find the critical points:

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

Domain of

x^{3} + 3 x^{2} + 4, - 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} + 4 \Rightarrow y = 4

Minimum (- 3, 4)

Plug the extreme point

x = - 2

into

x^{3} + 3 x^{2} + 4 \Rightarrow y = 8

Maximum (- 2, 8)

Plug the extreme point

x = 0

into

x^{3} + 3 x^{2} + 4 \Rightarrow y = 4

Minimum (0, 4)

Plug the extreme point

x = 2

into

x^{3} + 3 x^{2} + 4 \Rightarrow y = 24

Maximum (2, 24)

Minimum (- 3, 4), Maximum (- 2, 8), Minimum (0, 4), Maximum (2, 24)

f (x) = \frac{( x - 5 ) ^{2}}{x + 9}

f (x) = x^{5} - 2 x^{3} + 1

x^{4} - 8 x^{3} + 6

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

f (x) = - x + cos (3 π x)

What is the extreme f(x)=x^3+3x^2+4,-3<= x<= 2 ?
The extreme f(x)=x^3+3x^2+4,-3<= x<= 2 is Minimum(-3,4),Maximum(-2,8),Minimum(0,4),Maximum(2,24)