Question

求19题，跪谢

Answer 1

可导证明函数连续
当X=1时则aX+b=1
a=-1
则b=2

追问

如果连续，会怎么样

跪谢

Answer 2

(19)
f(x)= 2/(x^2+2) ; x≤1
= ax+b ; x>1

f(1-)=f(1)= 2/(1+2) =2/3
f(1+) = a+b
f(1-)=f(1+)=f(1)
a+b = 2/3 (1)

f'(1+)
=lim(h->1+) [ f(h) - f(1) ] /( h-1)
=lim(h->1+) [ 2/(h^2+2) - 2/3 ] /( h-1)
=lim(h->1+) [ 6 - 2(h^2+2) ]/ [3(h^2+2) .( h-1)]
=lim(h->1+) 2(1-h)(1+h)/ [3(h^2+2) .( h-1)]
=lim(h->1+) -2(1+h)/ [3(h^2+2) ]
=-2(1+1)/9
=-4/9

f'(1-) = f(1+)
=>
a = -4/9

from (1)
a+b = 2/3
-4/9 +b =2/3
b = 10/9