f(x)=ax^3+bx^2+cx f(x)在x=-1有极值曲线y=f(x)在(3,-24)处的切线方程为8x+y=0 求a,b,c
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![f(x)=ax^3+bx^2+cx f(x)在x=-1有极值曲线y=f(x)在(3,-24)处的切线方程为8x+y=0 求a,b,c](/uploads/image/z/7260750-54-0.jpg?t=f%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx+f%28x%29%E5%9C%A8x%3D-1%E6%9C%89%E6%9E%81%E5%80%BC%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%283%2C-24%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BA8x%2By%3D0+%E6%B1%82a%2Cb%2Cc)
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f(x)=ax^3+bx^2+cx f(x)在x=-1有极值曲线y=f(x)在(3,-24)处的切线方程为8x+y=0 求a,b,c
f(x)=ax^3+bx^2+cx f(x)在x=-1有极值曲线y=f(x)在(3,-24)处的切线方程为8x+y=0 求a,b,c
f(x)=ax^3+bx^2+cx f(x)在x=-1有极值曲线y=f(x)在(3,-24)处的切线方程为8x+y=0 求a,b,c
y'=3ax^2+2bx+c
将x=-1,y'=0代入得方程1 3a-2b+c=0
将x=3代入,得方程2 k=y'=27a+6b+c=-8
将x=3,y=-24代入得方程3 27a+9b+3c=-24
解这三个方程可得a,b,c
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