Question

第七题，求解答

Answer 1

f(x)
=x+1 ; x<0
=x ; x≥0
F(x) = ∫(-1->x) f(t) dt
case 1: -1≤x<0
F(x)
= ∫(-1->x) f(t) dt
= ∫(-1->x) (t+1) dt
=[ (1/2)t^2 +t ]|(-1->x)
=(1/2)x^2 +x + 1/2
case 2: 0≤x≤1
F(x)
= ∫(-1->x) f(t) dt
= ∫(-1->0) f(t) dt + ∫(0->x) f(t) dt
=1/2 +∫(0->x) t dt
=1/2 + (1/2)x^2
----------------
F(0-) = lim(x->0) [(1/2)x^2 +x + 1/2] = 1/2
F(0+)=F(0) =1/2 + 0 =1/2
x=0, F(x) 连续
=> F(x) 在 (-1,1) 连续
--------
F'(0-)
=lim(h->0) [ (1/2)h^2 +h + 1/2- F(0) ]/h
=lim(h->0) [(1/2)h + 1 ]
=1
F'(0+)
=lim(h->0) [1/2 + (1/2)h^2 - F(0) ]/h
=0
F'(0+) ≠ F'(0-)
F'(0) , 不存在
=> F(x) 在 (-1,0)∩(0,1) 可导