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受欢迎的函数 >

extreme f(x)=x+e^{-3x},-3<= x<= 4

解答

极值点

f (x) = x + e^{- 3 x}, - 3 \leq x \leq 4

解答

极大值 (- 3, - 3 + e^{9}), 极小值 (\frac{ln ( 3 )}{3}, \frac{ln ( 3 ) + 1}{3}), 极大值 (4, 4 + \frac{1}{e ^{12}})

求解步骤

找到临界点:

x = - 3, x = \frac{ln ( 3 )}{3}, x = 4

x + e^{- 3 x}, - 3 \leq x \leq 4

的定义域

: - 3 \leq x \leq 4

Find intervals:

递减

: - 3 < x < \frac{ln ( 3 )}{3},

递增

: \frac{ln ( 3 )}{3} < x < 4

将极值点

x = - 3

代入

x + e^{- 3 x} \Rightarrow y = - 3 + e^{9}

极大值 (- 3, - 3 + e^{9})

将极值点

x = \frac{ln ( 3 )}{3}

代入

x + e^{- 3 x} \Rightarrow y = \frac{ln ( 3 ) + 1}{3}

极小值 (\frac{ln ( 3 )}{3}, \frac{ln ( 3 ) + 1}{3})

将极值点

x = 4

代入

x + e^{- 3 x} \Rightarrow y = 4 + \frac{1}{e ^{12}}

极大值 (4, 4 + \frac{1}{e ^{12}})

极大值 (- 3, - 3 + e^{9}), 极小值 (\frac{ln ( 3 )}{3}, \frac{ln ( 3 ) + 1}{3}), 极大值 (4, 4 + \frac{1}{e ^{12}})

作图

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