Question

高数，导数问题

求解求解

Answer 1

函数在某一点可导，则此函数在这一点连续，且它的左右导数相等

Answer 2

培训要钱么，

Answer 3

可导则连续。则在x=1点，左右极限相等且等于在X=1处值，即
2/(1+1)=a+b
又因为可到，则左右导数相等
则对函数求导
-2x/(1+x^2)^2 |x=1 = a
则 a=-1/2, b=5/2

追答

修正：b=3/2

Answer 4

f(x)
=2/(1+x^2) ; x≤1
=ax+b ; x>1
f(1-)=f(1)= lim(x->1) [ 1/(1+x^2)] = 1/2
f(1+)=lim(x->1) ( ax+b) = a+b
=>
a+b = 1/2 (1)
f'(1-)
=lim(h->0) [1/[1+ (1+h)^2] - f(1) ] / h
=lim(h->0) [1/[1+ (1+h)^2] - 1/2 ] / h
=lim(h->0) [2 -(h^2+2h+2) ]/ [2h(h^2+2h+2)]
=lim(h->0) (-h^2-2h)/ [2h(h^2+2h+2)]
=lim(h->0) (-h-2)/ [2(h^2+2h+2)]
= -2/4
=-1/2
f'(1+)
=lim(h->0) [a(1+h)+b - f(1) ] / h
=lim(h->0) [a(1+h)+b - 1/2 ] / h
=lim(h->0) [ah+ (b+a - 1/2) ] / h
=lim(h->0) ah / h
=a
=> a = -1/2
from (1)
a+b = 1/2
-1/2 +b =1/2
b=1
(a,b) =( -1/2 , 1)