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受欢迎的函数 >

extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1

解答

极值点

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1

解答

极大值 (0, 12), 极小值 (\frac{1}{2}, \frac{19}{4}), 极大值 (1, 17)

求解步骤

找到临界点:

x = 0, x = \frac{1}{2}, x = 1

20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1

的定义域

: 0 \leq x \leq 1

Find intervals:

递减

: 0 < x < \frac{1}{2},

递增

: \frac{1}{2} < x < 1

将极值点

x = 0

代入

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = 12

极大值 (0, 12)

将极值点

x = \frac{1}{2}

代入

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = \frac{19}{4}

极小值 (\frac{1}{2}, \frac{19}{4})

将极值点

x = 1

代入

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = 17

极大值 (1, 17)

极大值 (0, 12), 极小值 (\frac{1}{2}, \frac{19}{4}), 极大值 (1, 17)

作图

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