设f(x)在(－∞,＋∞)内可导,且F(x)=f(x^2-1)+f(1-x^2),证明F'(1)=F'(－1)=lim(t→0)[f[(-1-t)^2-1]+f[1-(-1-t)^2]-2f(0)]/(-t)=lim(t→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/(-t)=-lim(t→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/t(令x=t)=-lim(x→0)[

Question

设f(x)在(－∞,＋∞)内可导,且F(x)=f(x^2-1)+f(1-x^2),证明F'(1)=F'(－1)
=lim(t→0)[f[(-1-t)^2-1]+f[1-(-1-t)^2]-2f(0)]/(-t)=lim(t→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/(-t)
=-lim(t→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/t (令x=t)
=-lim(x→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/x
=-F(1),关于这里不太懂

Answer 1

根据定义,F'(1)
= lim(dx→0)[ f[(1+dx)^2-1]+f[1-(1+dx)^2]-2f(0)]/dx
(1+dx)^2 = (-1-dx)^2,所以
= lim(dx→0)[ f[(-1-dx)^2-1]+f[1-(-1-dx)^2]-2f(0)]/dx
令t=-dx,上式可转化为
= lim(-t→0) [f(-1+t)^2 -1] + f[1-(-1+t)^2]-2f(0)]/(-t)
= - lim(t→0) [f(-1+t)^2 -1] + f[1-(-1+t)^2]-2f(0)]/t
根据定义,= - F'(-1)
楼主上面那个解法的最后一步,其实就是倒数的定义.
因为F'(1)根据倒数的定义
= lim(x→0) [F(1+x) - F(1)]/x
= lim(x→0)[f[(1+x)^2-1]+f[1-(1+x)^2]-2f(0)]/x