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受欢迎的函数 >

extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4

解答

极值点

f (x) = (x^{2} - 16)^{\frac{1}{9}}, - 5 \leq x \leq 4

解答

极大值 (- 5, 9^{\frac{1}{9}}), 极小值 (0, - 1 6^{\frac{1}{9}}), 极大值 (4, 0)

求解步骤

找到临界点:

x = - 5, x = - 4, x = 0, x = 4

(x^{2} - 16)^{\frac{1}{9}}, - 5 \leq x \leq 4

的定义域

: - 5 \leq x \leq 4

Find intervals:

递减

: - 5 < x < - 4,

递减

: - 4 < x < 0,

递增

: 0 < x < 4

将极值点

x = - 5

代入

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = 9^{\frac{1}{9}}

极大值 (- 5, 9^{\frac{1}{9}})

将极值点

x = 0

代入

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = - 1 6^{\frac{1}{9}}

极小值 (0, - 1 6^{\frac{1}{9}})

将极值点

x = 4

代入

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = 0

极大值 (4, 0)

极大值 (- 5, 9^{\frac{1}{9}}), 极小值 (0, - 1 6^{\frac{1}{9}}), 极大值 (4, 0)

作图

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