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受欢迎的函数 >

extreme f(x)=6sqrt(x^2+1)-x,0<= x<= 3

解答

极值点

f (x) = 6 x^{2} + 1 - x, 0 \leq x \leq 3

解答

极大值 (0, 6), 极小值 (\frac{1}{35}, 35), 极大值 (3, 610 - 3)

求解步骤

找到临界点:

x = 0, x = \frac{1}{35}, x = 3

6 x^{2} + 1 - x, 0 \leq x \leq 3

的定义域

: 0 \leq x \leq 3

Find intervals:

递减

: 0 < x < \frac{1}{35},

递增

: \frac{1}{35} < x < 3

将极值点

x = 0

代入

6 x^{2} + 1 - x \Rightarrow y = 6

极大值 (0, 6)

将极值点

x = \frac{1}{35}

代入

6 x^{2} + 1 - x \Rightarrow y = 35

极小值 (\frac{1}{35}, 35)

将极值点

x = 3

代入

6 x^{2} + 1 - x \Rightarrow y = 610 - 3

极大值 (3, 610 - 3)

极大值 (0, 6), 极小值 (\frac{1}{35}, 35), 极大值 (3, 610 - 3)

作图

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