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受欢迎的函数 >

extreme f(x)=3x^3-36x^2+108x+4,0<= x<= 9

解答

极值点

f (x) = 3 x^{3} - 36 x^{2} + 108 x + 4, 0 \leq x \leq 9

解答

极小值 (0, 4), 极大值 (2, 100), 极小值 (6, 4), 极大值 (9, 247)

求解步骤

找到临界点:

x = 0, x = 2, x = 6, x = 9

3 x^{3} - 36 x^{2} + 108 x + 4, 0 \leq x \leq 9

的定义域

: 0 \leq x \leq 9

Find intervals:

递增

: 0 < x < 2,

递减

: 2 < x < 6,

递增

: 6 < x < 9

将极值点

x = 0

代入

3 x^{3} - 36 x^{2} + 108 x + 4 \Rightarrow y = 4

极小值 (0, 4)

将极值点

x = 2

代入

3 x^{3} - 36 x^{2} + 108 x + 4 \Rightarrow y = 100

极大值 (2, 100)

将极值点

x = 6

代入

3 x^{3} - 36 x^{2} + 108 x + 4 \Rightarrow y = 4

极小值 (6, 4)

将极值点

x = 9

代入

3 x^{3} - 36 x^{2} + 108 x + 4 \Rightarrow y = 247

极大值 (9, 247)

极小值 (0, 4), 极大值 (2, 100), 极小值 (6, 4), 极大值 (9, 247)

作图

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