1)若f(x)=ax+b,则 f[(x1+x2)/2]=[f(x1)+f(x2)]/2;2)若f(x)=x^2+ax+b,则 f[(x1+x2)/2]≤[f(x1)+f(x2)]/2请写过程.谢
热心网友
1)f(x)=ax+b---f((x1+x2)/2=a(x1+x2)/2+b& [f(x1)+f(x2)]/2=[(ax1+b)+(ax2+b)]/2=a(x1+x2)/2+b---f((x1+x2)/2)=[f(x1)+f(x2)]/2.2)f(x)=x^2+ax+b---f((x1+x2)/2)=(x1+x2)^2/4+a(x1+x2)/2+b& [f(x1)+f(x2)]/2=[(x1^2+ax1+b)+(x2+ax2+b)]/2=(x1^2+x2^2)/2+a(x1+x2)/2+b因为(x1+x2)^2/4-(x1+x2)^2/2=(x1^2+x2^2+2x1x2)/4-2(x1^2+x2^2)/4=(2x1x2-x1^2+x2^2)/4=-(x1-x2)^2/4=f((x1+x2)/2)=<[f(x1)+f(x2)]成立。
热心网友
1) f[(x1+x2)/2]=[(ax1+ax2)/2+b]=(ax1+ax2+2b)/2=(ax1+b)/2+(ax2+b)/2=]=[f(x1)+f(x2)]/22)f[(x1+x2)/2]=[(x1+x2)/2]^2+a(x1+x2)/2+b=[(x1/2)^2+ax1/2+b]+[(x2/2)^2+ax2/2}<==[(x1/2)^2+ax1/2+b]+[(x2/2)^2+ax2/2+b]=[f(x1)+f(x2)]/2