设函数f(x)=e^x*(ax^2+x+1),切曲线y=f(x)在x=1处的切线方程与x轴平行1.求a的值,并讨论f(x)的单调性2.证明:当θ∈[0,π/2]时,︱f(cosθ)-f(sinθ)︱

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设函数f(x)=e^x*(ax^2+x+1),切曲线y=f(x)在x=1处的切线方程与x轴平行1.求a的值,并讨论f(x)的单调性2.证明:当θ∈[0,π/2]时,︱f(cosθ)-f(sinθ)︱
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设函数f(x)=e^x*(ax^2+x+1),切曲线y=f(x)在x=1处的切线方程与x轴平行1.求a的值,并讨论f(x)的单调性2.证明:当θ∈[0,π/2]时,︱f(cosθ)-f(sinθ)︱
设函数f(x)=e^x*(ax^2+x+1),切曲线y=f(x)在x=1处的切线方程与x轴平行
1.求a的值,并讨论f(x)的单调性2.证明:当θ∈[0,π/2]时,︱f(cosθ)-f(sinθ)︱

设函数f(x)=e^x*(ax^2+x+1),切曲线y=f(x)在x=1处的切线方程与x轴平行1.求a的值,并讨论f(x)的单调性2.证明:当θ∈[0,π/2]时,︱f(cosθ)-f(sinθ)︱
1.对f(x)求导
f’(x)=(2ax+1)e^x+(ax^2+x+1)e^x=(ax^2+(2a+1)x+2)e^x
由f’(1)=0得3a+3=0 a=-1
所以令f'(x)=(-x^2-x+2)e^x=0 得x=-2或者x=1
在[-2,1],-x^2-x+2≥0 f(x)单调递增
(-∞,-2)∪(1,+∞)上f’(x)