设函数f（x）={x^2,x≤1;ax+b,x>1}为使函数f（x）在x=1处连续且可导,a、b应取什么值?

f(1)=1 linf(x)x→1+=a+bx≤1 f'(x)=2x limf'(x)x→1-=2x>1 f'(x)=a limf'(x)x→1+=a在x=1处连续 f(1)=linf(x)x→1+1=a+b.(1)在x=1处连续且可导limf'(x)x→1-=limf'(x)x→1+2=a.(2)解(1)(2)a=2,b=-1linf(x)x→1+表示...