What is the extreme f(x)=2x^3+3x^2+4,-2<= x<= 1 ?

The extreme f(x)=2x^3+3x^2+4,-2<= x<= 1 is Minimum(-2,0),Maximum(-1,5),Minimum(0,4),Maximum(1,9)

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extreme f(x)=2x^3+3x^2+4,-2<= x<= 1

extreme points

f (x) = 2 x^{3} + 3 x^{2} + 4, - 2 \leq x \leq 1

Minimum (- 2, 0), Maximum (- 1, 5), Minimum (0, 4), Maximum (1, 9)

Solution steps

Find the critical points:

x = - 2, x = - 1, x = 0, x = 1

Domain of

2 x^{3} + 3 x^{2} + 4, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

2 x^{3} + 3 x^{2} + 4 \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = - 1

into

2 x^{3} + 3 x^{2} + 4 \Rightarrow y = 5

Maximum (- 1, 5)

Plug the extreme point

x = 0

into

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

Minimum (0, 4)

Plug the extreme point

x = 1

into

2 x^{3} + 3 x^{2} + 4 \Rightarrow y = 9

Maximum (1, 9)

Minimum (- 2, 0), Maximum (- 1, 5), Minimum (0, 4), Maximum (1, 9)

f (x) = x^{3} + 3 x^{2} - 24 x [0.3]

f (x) = - \frac{2}{x - 7}

f (x) = x^{2} + x + 4

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

y = - 2 x^{2} + 6

What is the extreme f(x)=2x^3+3x^2+4,-2<= x<= 1 ?
The extreme f(x)=2x^3+3x^2+4,-2<= x<= 1 is Minimum(-2,0),Maximum(-1,5),Minimum(0,4),Maximum(1,9)