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Popular >

extreme f(x)=|x^3-3x^2+2|

Solution

extreme points

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

Solution

Minimum (1 - 3, 0), Maximum (0, 2), Minimum (1, 0), Maximum (2, 2), Minimum (1 + 3, 0)

Solution steps

Find the critical points:

x = 1 - 3, x = 0, x = 1, x = 2, x = 1 + 3

Find intervals:

Decreasing

: - \infty < x < 1 - 3,

Increasing

: 1 - 3 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < 1 + 3,

Increasing

: 1 + 3 < x < \infty

Plug the extreme point

x = 1 - 3

into

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

Minimum (1 - 3, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = 1

into

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

Minimum (1, 0)

Plug the extreme point

x = 2

into

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

Maximum (2, 2)

Plug the extreme point

x = 1 + 3

into

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

Minimum (1 + 3, 0)

Minimum (1 - 3, 0), Maximum (0, 2), Minimum (1, 0), Maximum (2, 2), Minimum (1 + 3, 0)

Graph

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