对于函数f(x)=lgx定义域中任意X1,X2(X1≠X2)有如下结论:①f(x1+x2)=f(x1)+f(x2);②f(x1·x2)=f(x1)+f(x2);③f(x1)-f(x2)/x1-x2>0;④f(x1+x2/2)
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![对于函数f(x)=lgx定义域中任意X1,X2(X1≠X2)有如下结论:①f(x1+x2)=f(x1)+f(x2);②f(x1·x2)=f(x1)+f(x2);③f(x1)-f(x2)/x1-x2>0;④f(x1+x2/2)](/uploads/image/z/2084064-24-4.jpg?t=%E5%AF%B9%E4%BA%8E%E5%87%BD%E6%95%B0f%28x%29%3Dlgx%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%AD%E4%BB%BB%E6%84%8FX1%2CX2%28X1%E2%89%A0X2%29%E6%9C%89%E5%A6%82%E4%B8%8B%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0f%28x1%2Bx2%29%3Df%28x1%29%2Bf%28x2%29%3B%E2%91%A1f%28x1%C2%B7x2%29%3Df%28x1%29%2Bf%28x2%29%3B%E2%91%A2f%28x1%29-f%28x2%29%EF%BC%8Fx1-x2%3E0%3B%E2%91%A3f%28x1%2Bx2%2F2%29)
对于函数f(x)=lgx定义域中任意X1,X2(X1≠X2)有如下结论:①f(x1+x2)=f(x1)+f(x2);②f(x1·x2)=f(x1)+f(x2);③f(x1)-f(x2)/x1-x2>0;④f(x1+x2/2)
对于函数f(x)=lgx定义域中任意X1,X2(X1≠X2)有如下结论
:①f(x1+x2)=f(x1)+f(x2);②f(x1·x2)=f(x1)+f(x2);③f(x1)-f(x2)/x1-x2>0;④f(x1+x2/2)
对于函数f(x)=lgx定义域中任意X1,X2(X1≠X2)有如下结论:①f(x1+x2)=f(x1)+f(x2);②f(x1·x2)=f(x1)+f(x2);③f(x1)-f(x2)/x1-x2>0;④f(x1+x2/2)
②正确
由对数运送法则
f(x1·x2)
=lg(x1x2)
=lgx1+lgx2
=f(x1)+f(x2);
③正确
f(x)=lgx定义域中是增函数,
0
①不正确
x1=10,x2=100
f(10+100)=lg110
④不正确
x1=10,x2=100
f(10+50)=lg60
只有一个(2)是对的,
lgx1+lgx2=lg(x1*x2),用这个式子去推断,(1)错
lgx1-lgx2=lg(x1/x2),用这个式子去推断,(3)错
lgx1+(lgx2)/2=lgx1+1/2*lgx2=lgx1+lg(x2)^1/2=lg(x1*√x2),用这个式子去推断,(4)错.
②③
由f(x1*x2)=lgx1*x2=lgx1+lgX2=f(x1)+f(x2)②正确
函数f(x)=lgx是增函数,所以f(x1)-f(x2)与x1-x2的符号相同③正确
因为函数f(x)=lgx是凸函数在函数图象上f((x1+x2)/2)高于(f(x1)+f(x2))/2,后一个可看作梯形中位线长。所以应是大于号④错误
d