# 微積分函數正解

### 1 個解答

• 麻辣
Lv 7
8 年前
最佳解答

(1) f(x)=x^3-4x^2-3x+1f'(x)=3x^2-8x-3 f'(0)=-3=(y-1)/x3x+y=1...........ans

(2) y=1/(3x^2+x+1)^(5/3)y'=-5*(3x+1)/[3(3x^2+x+1)^(8/3)]............ans

(3) f(x)=(x+2)^8x^6係數=C(8,2)*2^2=112................ans

(4) A=(0,-3), B=(-2,1), 中垂線=?中點M=(A+B)/2=(-1,-1)AB=from A to B=(-1,2)Normal=N(AB)=(2,1) slope=1/2=(y+1)/(x+1)x-2y=1................ans

(5) f(x)=(x+1)^4*(2x-1)/(x^2+2)f(x)*(x^2+2)=(x+1)^4*(2x-1)f'(x)*(x^2+2)+f(x)*2x=4(x+1)^3*(2x-1)+2*(x+1)^4f'(x)=[4*(x+1)^3*(2x-1)+2*(x+1)^4-2x*f(x)]/(x^2+2)=(x+1)^3*[4*(2x-1)+2*(x+1)]/(x^2+2)-2x*(x+1)^4*(2x-1)/(x^2+2)^2=[(x+1)^3*(10x-2)*(x^2+2)-2x*(x+1)^4*(2x-1)]/(x^2+2)^2=2*(x+1)^3*[(5x-1)*(x^2+2)-x*(x+1)(2x-1)]/(x^2+2)^2=2*(x+1)^3*(3x^3-2x^2+11x-2)/(x^2+2)^2............ans

(6) y^2=(x^2-2)/(x-1)2y*y'=[2x(x-1)-(x^2-2)]/(x-1)^2y'=[2x(x-1)-(x^2-2)]/[2y(x-1)^2]=(x^2-2x+2)/[2(x-1)^2]*[(x-1)/(x^2-2)]^0.5=(x^2-2x+2)/2[(x-1)^1.5*(x^2-2)^0.5]=(x^2-2x+2)/{2(x-1)*[(x-1)(x^2-2)]^0.5}

(7) y=(2x+1)^3/(x^2+1)y'=[6(x^2+1)(2x+1)^2-2x(2x+1)^3]/(x^2+1)^2=2(2x+1)^2*[3(x^2+1)-x(2x+1)]/(x^2+1)^2=2*(2x+1)^2*(x^2-x+3)/(x^2+1)^2............ans

(8) y=Limit<x->1>[2-(x+3)^0.5]/(x^2-1)x->1使分子母=0/0=不確定,可以使用l'Hopital定理:y=Limit<x->1>[2-(x+3)^0.5]'/(x^2-1)'=Limit<x->1>[-0.5*(x+3)^-0.5]/2x=-0.5*0.5/2=-1/8.................ans

(9) f(x)=(x^2+5)/(x^2-1)a=Limit<x->±∞>f(x)=1b=Limit<x->1>f(x)=∞(x<1,x>1)c=Limit<x->1>f(x)=-∞(1<x<1)d=f(5)=0

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