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受欢迎的函数 >

extreme f(x)=(5x)/(x^2+16),0<= x<= 12

解答

极值点

f (x) = \frac{5 x}{x ^{2} + 16}, 0 \leq x \leq 12

解答

极小值 (0, 0), 极大值 (4, \frac{5}{8}), 极小值 (12, \frac{3}{8})

求解步骤

找到临界点:

x = 0, x = 4, x = 12

\frac{5 x}{x ^{2} + 16}, 0 \leq x \leq 12

的定义域

: 0 \leq x \leq 12

Find intervals:

递增

: 0 < x < 4,

递减

: 4 < x < 12

将极值点

x = 0

代入

\frac{5 x}{x ^{2} + 16} \Rightarrow y = 0

极小值 (0, 0)

将极值点

x = 4

代入

\frac{5 x}{x ^{2} + 16} \Rightarrow y = \frac{5}{8}

极大值 (4, \frac{5}{8})

将极值点

x = 12

代入

\frac{5 x}{x ^{2} + 16} \Rightarrow y = \frac{3}{8}

极小值 (12, \frac{3}{8})

极小值 (0, 0), 极大值 (4, \frac{5}{8}), 极小值 (12, \frac{3}{8})

作图

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