求大神指点,f(|x|)在点x=0处可导的充分必要条件为什么是f'(0)=0?
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证明:①充分性: f(0)=0 ,则:
F'(0)
=lim(x->0) [f(x)(1+|sinx|)-f(0)(1+|sin0|]/x
=lim(x->0) f(x)(1+|sinx|)/x
=lim(x->0) [f(x)-f(0)]/x* (1+|sinx|)
= f'(0)*1
= f'(0)
②必要性:F(x)在x=0处可导,则:
F'(0+0)=F'(0-0)
由导数极限定理【此处也可改为极限式计算】:
F'(0+0)
=lim(x->0+) [f(x)(1+sinx)]'
=lim(x->0+) [f'(x)(1+sinx)+f(x)*cosx]
=f'(0)+f(0)
F'(0-0)
=lim(x->0+) [f(x)(1-sinx)]'
=lim(x->0+) [f'(x)(1-sinx)-f(x)*cosx]
=f'(0)-f(0)
∵F'(0+0)=F'(0-0)
∴f'(0)+f(0)=f'(0)-f(0)
∴ f(0)=0
希望对你有所帮助 还望采纳~~
F'(0)
=lim(x->0) [f(x)(1+|sinx|)-f(0)(1+|sin0|]/x
=lim(x->0) f(x)(1+|sinx|)/x
=lim(x->0) [f(x)-f(0)]/x* (1+|sinx|)
= f'(0)*1
= f'(0)
②必要性:F(x)在x=0处可导,则:
F'(0+0)=F'(0-0)
由导数极限定理【此处也可改为极限式计算】:
F'(0+0)
=lim(x->0+) [f(x)(1+sinx)]'
=lim(x->0+) [f'(x)(1+sinx)+f(x)*cosx]
=f'(0)+f(0)
F'(0-0)
=lim(x->0+) [f(x)(1-sinx)]'
=lim(x->0+) [f'(x)(1-sinx)-f(x)*cosx]
=f'(0)-f(0)
∵F'(0+0)=F'(0-0)
∴f'(0)+f(0)=f'(0)-f(0)
∴ f(0)=0
希望对你有所帮助 还望采纳~~
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