设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4 1.设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和41.当a=3时,求曲线y=f(x)在点(1,f(x))处的切线
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![设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4 1.设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和41.当a=3时,求曲线y=f(x)在点(1,f(x))处的切线](/uploads/image/z/5254484-68-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%EF%BC%88a%2F3%EF%BC%89x%5E3%2Bbx%5E2%2Bcx%EF%BC%88a%EF%BC%9E0%EF%BC%89%2C%E6%96%B9%E7%A8%8Bf%EF%BC%87%EF%BC%88x%EF%BC%89%EF%B9%A39x%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E6%98%AF1%E5%92%8C4+1.%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%EF%BC%88a%2F3%EF%BC%89x%5E3%2Bbx%5E2%2Bcx%EF%BC%88a%EF%BC%9E0%EF%BC%89%2C%E6%96%B9%E7%A8%8Bf%EF%BC%87%EF%BC%88x%EF%BC%89%EF%B9%A39x%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E6%98%AF1%E5%92%8C41.%E5%BD%93a%3D3%E6%97%B6%2C%E6%B1%82%E6%9B%B2%E7%BA%BFy%3Df%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%88x%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF)
设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4 1.设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和41.当a=3时,求曲线y=f(x)在点(1,f(x))处的切线
设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4 1.
设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4
1.当a=3时,求曲线y=f(x)在点(1,f(x))处的切线方程
设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和4 1.设函数f(x)=(a/3)x^3+bx^2+cx(a>0),方程f'(x)﹣9x=0的两根是1和41.当a=3时,求曲线y=f(x)在点(1,f(x))处的切线
c/a=4,(9-2b)/a=5;所以:a=3时,c=12,b=-3.f(x)=x^3-3x^2+12x,f(1)=10,
f'(x)=3x^2-6x+12,
f'(1)=9
所以切线方程为:y-10=9(x-1),
y=9x+1
a=3时,f(x)=x^3+bx^2+cx f'(x)-9x=(3x^2+2bx+c)-9x => f'(x)-9x=3x^2+(2b-9)x+c
∴ 2b-9=-(1+41)*3 => b=-117/2
c=1*41*3=123
∴f(x)=x^3-(117x^2)/2+123x f(1)=121/2 f'(1)...
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a=3时,f(x)=x^3+bx^2+cx f'(x)-9x=(3x^2+2bx+c)-9x => f'(x)-9x=3x^2+(2b-9)x+c
∴ 2b-9=-(1+41)*3 => b=-117/2
c=1*41*3=123
∴f(x)=x^3-(117x^2)/2+123x f(1)=121/2 f'(1)= 9 即切线斜率 k=9
∴点(1,121/2)处的切线方程为: y-121/2=9(x-1)
=> 18x-2y+103=0
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y'=ax^2+2bx+c,方程f'(x)﹣9x=0的两根是1和41,