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/3768955-43-5.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
f'=ax²+2bx+c
f'(x)﹣9x=0为ax²-(9-2b)x+c=0根是1和4
a-9+2b+c=0
16a-36+8b+c=0
c=4a
b=9/2-5a/2
f'=3x²-6x+12
当x=1时f'=9
y=f(x)在点(1,f(x))处的切线方程为
y-10=9(x-1)
9x-y+1=0
f'(x)=ax²+2bx+c
f'(x)-9x=ax²+(2b-9)x+c=0的两根为1和4
由韦达定理得:
(9-2b)/a=1+4=5
c/a=1*4=4
a=3
解得:
a=3,b=-3,c=12
∴f(x)=x³-3x²+12x
f(1)=1-3+12=10
f'(x)...
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f'(x)=ax²+2bx+c
f'(x)-9x=ax²+(2b-9)x+c=0的两根为1和4
由韦达定理得:
(9-2b)/a=1+4=5
c/a=1*4=4
a=3
解得:
a=3,b=-3,c=12
∴f(x)=x³-3x²+12x
f(1)=1-3+12=10
f'(x)=3x²-6x+12
f'(1)=3-6+12=9
所以切线方程为:
y-10=9(x-1)
即y=9x+1
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