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受欢迎的函数 >

extreme f(x)=x^3-5x^2+3x+10,-2<= x<= 4

解答

极值点

f (x) = x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4

解答

极小值 (- 2, - 24), 极大值 (\frac{1}{3}, \frac{283}{27}), 极小值 (3, 1), 极大值 (4, 6)

求解步骤

找到临界点:

x = - 2, x = \frac{1}{3}, x = 3, x = 4

x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4

的定义域

: - 2 \leq x \leq 4

Find intervals:

递增

: - 2 < x < \frac{1}{3},

递减

: \frac{1}{3} < x < 3,

递增

: 3 < x < 4

将极值点

x = - 2

代入

x^{3} - 5 x^{2} + 3 x + 10 \Rightarrow y = - 24

极小值 (- 2, - 24)

将极值点

x = \frac{1}{3}

代入

x^{3} - 5 x^{2} + 3 x + 10 \Rightarrow y = \frac{283}{27}

极大值 (\frac{1}{3}, \frac{283}{27})

将极值点

x = 3

代入

x^{3} - 5 x^{2} + 3 x + 10 \Rightarrow y = 1

极小值 (3, 1)

将极值点

x = 4

代入

x^{3} - 5 x^{2} + 3 x + 10 \Rightarrow y = 6

极大值 (4, 6)

极小值 (- 2, - 24), 极大值 (\frac{1}{3}, \frac{283}{27}), 极小值 (3, 1), 极大值 (4, 6)

作图

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