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受欢迎的函数 >

extreme f(x)=x^2-x-2,0<= x<= 2

解答

极值点

f (x) = x^{2} - x - 2, 0 \leq x \leq 2

解答

极大值 (0, - 2), 极小值 (\frac{1}{2}, - \frac{9}{4}), 极大值 (2, 0)

求解步骤

找到临界点:

x = 0, x = \frac{1}{2}, x = 2

x^{2} - x - 2, 0 \leq x \leq 2

的定义域

: 0 \leq x \leq 2

Find intervals:

递减

: 0 < x < \frac{1}{2},

递增

: \frac{1}{2} < x < 2

将极值点

x = 0

代入

x^{2} - x - 2 \Rightarrow y = - 2

极大值 (0, - 2)

将极值点

x = \frac{1}{2}

代入

x^{2} - x - 2 \Rightarrow y = - \frac{9}{4}

极小值 (\frac{1}{2}, - \frac{9}{4})

将极值点

x = 2

代入

x^{2} - x - 2 \Rightarrow y = 0

极大值 (2, 0)

极大值 (0, - 2), 极小值 (\frac{1}{2}, - \frac{9}{4}), 极大值 (2, 0)

作图

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