f(x)=[1-√1-x]/x,x≠0处处可导,求a=b=f(x)=ax+b,x≧0

Question

f(x)=[1-√1-x]/x,x≠0 处处可导,求a= b= f(x)=ax+b,x≧0

Answer 1

已知f(x)=[1-√(1-x)]/x,当x≠0 ；f(x)=ax+b,当x≧0；处处可导,则a=,b=
可导必连续.x→0limf(x)=x→0lim[1-√(1-x)]/x=x→0lim[1-√(1-x)][1+√(1-x)]/x[1+√(1-x)]
=x→0lim 1/[1+√(1-x)]=1/2=f(0)=b,即b=1/2；
(f(x)在x=0处连续,说明f(x)在x=0处有定义,有极限,且极限等于定义,即x→0lim f(x)=f(0))
f(x)=[1-√(1-x)]/x=(1/x)-[√(1-x)]/x
f′(x)=-(1/x²)-{-x/[2√(1-x)]-√(1-x)}/x²=-(1/x²)+[x+2(1-x)]/[2x²√(1-x)]=-(1/x²)+(2-x)/[2x²√(1-x)]
=[-2√(1-x)+2-x]/[2x²√(1-x)]=[-2+(2-x)/√(1-x)]/(2x²)
x→0lim f′(x)=x→0lim [-2+(2-x)/√(1-x)]/(2x²)(0/0型)
=x→0lim [-√(1-x)+(2-x)/2√(1-x)]/4x(此处用了罗比塔法则)
=x→0lim [-2(1-x)+(2-x)]/[8x√(1-x)](分子通分并整理)
=x→0lim{ x/[8x√(1-x)]}=x→0lim1/[8√(1-x)]=1/8=a
即a=1/8,b=1/2.
(f(x)在x=0处可导,同连续的含意一样,说明x→0lim f′(x)=f′(0))