已知f(x)=x的平方+2x+3分之x(x属于2,到正无穷)证明f(x)是增函数,并求最小值
已知f(x)=x的平方+2x+3分之x(x属于2,到正无穷)证明f(x)是增函数,并求最小值
f(x)=x的平方+2x+3/x
设2≤x1分母呢?你的输入,没看懂,若有分式得话应该用括号f(x)=x/(x²+2x+3) f(x1)-f(x2)=x1/(x²1+2x1+3)-x2(x²2+x2+3)=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x1x²2-x²1x2)+3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)] =[x1x2(x1-x2)-3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x1-x2)(x1x2-3)]/[(x²1+2x1+3)(x²2+x2+3)] ∵ 2≤x1