求函数fx＝x²ex.的单调区间与极值 函数fx=e∧x-ex的单调区间和极值
\u6c42\u51fd\u6570fx=x-\u6839\u53f7x\u7684\u5355\u8c03\u533a\u95f4\u4e0e\u6781\u503c\u3002\u8981\u8fc7\u7a0b.f(x) = x - \u221ax\uff0c\uff08x \u2265 0\uff09 \uff0c
f ' (x) = 1 - 1/2\u221ax \uff0c
\u4ee4 f ' (x) > 0 \u5f97 x > 1/4 \uff0c\u4ee4 f ' (x) < 0 \u5f97 0 < x < 1/4 \uff0c
\u6240\u4ee5\u51fd\u6570\u5728\uff080\uff0c1/4\uff09\u4e0a\u51cf\uff0c\u5728\uff081/4\uff0c+\u221e\uff09\u4e0a\u589e\uff0c
\u51fd\u6570\u5728 x = 1/4 \u5904\u53d6\u6781\u5c0f\u503c - 1/4 \u3002
f(x)=x²eˣ 定义域x∈R
f'(x)=2xeˣ+x²eˣ=eˣ(x²+2x)
驻点x₁=-2 x₂=0
f''(x)=eˣ(x²+2x)+eˣ(2x+2)=eˣ(x²+4x+2)
f''(-2)<0 x₁=-2是极大值点，极大值f(-2)=4/e²
f''(0)>0, x₂=0是极小值点,极小值f(0)=0
单调递增区间x∈(-∞,-2)∪(0,+∞)
单调递减区间x∈(-2,0)
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