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

The extreme f(x)=2x^3+3x^2,1<= x<= 2 is Minimum(1,5),Maximum(2,28)

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

extreme points

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

Minimum (1, 5), Maximum (2, 28)

Solution steps

Find the critical points:

x = 1, x = 2

Domain of

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

Find intervals:

Increasing

: 1 < x < 2

Plug the extreme point

x = 1

into

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

Minimum (1, 5)

Plug the extreme point

x = 2

into

2 x^{3} + 3 x^{2} \Rightarrow y = 28

Maximum (2, 28)

Minimum (1, 5), Maximum (2, 28)

y = x^{3} - 8 x^{2} - 12 x + 8

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

f (x, y) = (2 x - x^{2}) (2 y - y^{2})

f (x) = (x - 2) e^{x} + 2

e^{7 x} + e^{- x}

What is the extreme f(x)=2x^3+3x^2,1<= x<= 2 ?
The extreme f(x)=2x^3+3x^2,1<= x<= 2 is Minimum(1,5),Maximum(2,28)