Question

两道数学题 急

若f(x)=ax+b 证明f[(x1+x20/2]=[f(x1)+f(x2)]/2
若g(x)=x平方+ax+b 证明g[(x1+x2)/2]小于等于[g(x1)+g(x2)]/2
过程重要..过程...

Answer 1

f((x1+x2)/2)=a(x1+x2)/2+b
=ax1/2+ax2/2+b/2+b/2
=(ax1+b)/2+(ax2+b)/2
=((ax1+b)+(ax2+b))/2
=(f(x1)+f(x2))/2
g((x1+x2)/2)-(g(x1)+g(x2))/2
=(((x1+x2)/2)^2+a((x1+x2)/2)+b)-((x1^2+ax1+b)+(X2^2+ax2+b))/2
=((x1^2+x2^2+2x1x2)/4+a(x1+x2))/2+b)-(x1^2+x2^2+a(x1+x2)+2b)/2
=-(x1^2+x2^2-2x1x2)/4
=-(x1-x2)^2/4<=0
所以g[(x1+x2)/2]小于等于[g(x1)+g(x2)]/2