Question

已知f(x)=(x-a)的绝对值,g(x)=x^2+2ax+1(a>0),且满足f(0)=g(0).

Answer 1

f(0)=|0-a|=|a|=a

g(0)=0+0+1=1

因为f(0)=g(0),所以a=1

f(x)+g(x)=|x-1|+x²+2x+1

（1）当x≥1时,f(x)+g(x)=(x-1)+x²+2x+1=x²+3x

x越大,x²+3x越大,所以[1,+∞)是f(x)+g(x)的单调增区间

（2）当x