求曲线y=1 3x3-3 2x2 2x在(0,0)切线
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设切点为(x0,y0),∵y′=3x2-6x+2,∴切线斜率为3x02-6x0+2,∴切线方程为y-y0=(3x02-6x0+2)(x-x0)∵切点在曲线C上,∴y0=x03-3x02+2x0-1,①
第二项应该是3x^2吧,先求导设k=3x^2+6x+6=3(x+1)^2+3当x=-1时斜率最小为3,切点为(-1,-14),所以切线方程为3x-y-11=0
先求在点M的导数原含数y=2x-x3则导含数y=2-3x2M处的斜率是k=-1所以切线方程为y+1=(-1)(x+1)
y`=2-3x^2x=1y`=-1y-1=-1(x-1)y-1=1-xx+y=2
∵f(x)=x3+bx2+cx+d,由图象知,-1+b-c+d=0,0+0+0+d=0,8+4b+2c+d=0,∴d=0,b=-1,c=-2∴f′(x)=3x2+2bx+c=3x2-2x-2.由题意有
y=1/3x3+4/3y的导数y'=x²,所以x=a处的斜率为a²
y'=6x^2切线斜率k=y'|(x=1)=6又切点为(1,3)∴切线方程为y-3=6(x-1)即6x-y-3=0
k=3*1-2=1,切线方程是:x-y-1=0
由题意,得斜率=3×1平方=3所以切线方程为y-1=3(x-1)即y=3x-2
x=2,则y=2所以切点(2,2)y'=3x²-2x=2,y'=10即切线斜率是10所以是10x-y-18=0
f'(x)=3x^2f'(1)=3由点斜式得切线方程:y=3(x-1)+2=3x-1
求二阶导数为0的点,即6x+6=0,(-1,-9);因为在(-1)时相反.此处斜率最小.切线方程为3x-y-6=0
设切点为(x0,y0)根据题意得y'=3x²∴k=y'|x=x0=3x0²∴切线为y-1=(3x0²)(x-1)①又∵切点在曲线上∴y0=x0³②由①②得x0&
对函数y=1/3x3+4/3求导可得y′=x^2所以,曲线在点P(2,4)处的斜率是:k=y′|x=2=4因此,曲线上点P(2,4)处的切线方程是:y-4=4(x-2)整理得:4x-y-4=0
x^3+y^3+(x+1)cos(πy)+9=0x=-1时,-1+y^3-0+9=0y^3=-8y=-23x^2+3y^2y'+(x+1)π(-sin(πy))y'+cos(πy)=03x^2+(3y
y'=3x²+1令x=2,y'=13所以在(2,-6)处的切线斜率为13,所以切线方程:y+6=13(x-2)即:y=13x-32
答:y=x^3求导:y'(x)=3x^2x=1时:y(1)=1,y'(1)=3切点(1,1),切线斜率k=3切线方程y-1=3(x-1)所以:切线为y=3x-2
y=x³-2x+4则:y'=3x²-2则切线斜率是:k=y'(x=1)=1切点是(1,3)则切线是:x-y+2=0
y′=3x2-1≥-1,∴tanα≥-1,∴[0,π2)∪[3π4,π),故答案为[0,π2)∪[3π4,π)
y′=12x3-12x2,y″=36x2-24x=12x(3x-2)令y″=0解得,x=0或x=23.所以曲线的拐点为(0,1),(23,1127).当x<0或x>23时,y″>0,则曲线的凹区间为(