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受欢迎的函数 >

extreme f(x)=3x^3-30x^2+75x+6,0<= x<= 9

解答

极值点

f (x) = 3 x^{3} - 30 x^{2} + 75 x + 6, 0 \leq x \leq 9

解答

极小值 (0, 6), 极大值 (\frac{5}{3}, \frac{554}{9}), 极小值 (5, 6), 极大值 (9, 438)

求解步骤

找到临界点:

x = 0, x = \frac{5}{3}, x = 5, x = 9

3 x^{3} - 30 x^{2} + 75 x + 6, 0 \leq x \leq 9

的定义域

: 0 \leq x \leq 9

Find intervals:

递增

: 0 < x < \frac{5}{3},

递减

: \frac{5}{3} < x < 5,

递增

: 5 < x < 9

将极值点

x = 0

代入

3 x^{3} - 30 x^{2} + 75 x + 6 \Rightarrow y = 6

极小值 (0, 6)

将极值点

x = \frac{5}{3}

代入

3 x^{3} - 30 x^{2} + 75 x + 6 \Rightarrow y = \frac{554}{9}

极大值 (\frac{5}{3}, \frac{554}{9})

将极值点

x = 5

代入

3 x^{3} - 30 x^{2} + 75 x + 6 \Rightarrow y = 6

极小值 (5, 6)

将极值点

x = 9

代入

3 x^{3} - 30 x^{2} + 75 x + 6 \Rightarrow y = 438

极大值 (9, 438)

极小值 (0, 6), 极大值 (\frac{5}{3}, \frac{554}{9}), 极小值 (5, 6), 极大值 (9, 438)

作图

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