What is the extreme f(x)=x^3+2x^2-7x,0<= x<= 2 ?

The extreme f(x)=x^3+2x^2-7x,0<= x<= 2 is Maximum(0,0),Minimum(1,-4),Maximum(2,2)

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extreme f(x)=x^3+2x^2-7x,0<= x<= 2

extreme points

f (x) = x^{3} + 2 x^{2} - 7 x, 0 \leq x \leq 2

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

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

x^{3} + 2 x^{2} - 7 x, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = - 4

Minimum (1, - 4)

Plug the extreme point

x = 2

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = 2

Maximum (2, 2)

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

f (x) = 2 x^{3} - 21 x^{2} + 72 x - 72.5

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

f (x, y) = 5 x^{2} + 5 y^{2} + 3

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

x + \frac{25}{x}

What is the extreme f(x)=x^3+2x^2-7x,0<= x<= 2 ?
The extreme f(x)=x^3+2x^2-7x,0<= x<= 2 is Maximum(0,0),Minimum(1,-4),Maximum(2,2)