已知函数f(x)=(x^(1/3)-x^(-1/3))/5,g(x)=(x^(1/3)+x^(-1/3))/5

来源：百度知道 编辑：UC知道 时间：2024/05/01 11:38:26
证明:f(x)是奇函数,并求f(x)的单调区间，2.分别计算f(4)-5f(2)g(2)和f(9)-5f(3)g(3)的值 ,由此概括出涉及函数f(x)和g(x)的对所有不等等于零的实数x都成立的一个等式加以证明。

(1)f(x)=(x^(1/3)-x^(-1/3))/5
g(x)=(x^(1/3)+x^(-1/3))/5
X属于R
则有:f(-x)=[(-x)^(1/3)-(-x)^(-1/3)]/5
=[x^(-1/3)-x^(1/3)]/5
=-[x^(1/3)-x^(-1/3)]/5
=-f(x)
g(-x)=[(-x)^(1/3)+(-x)^(-1/3)]/5
=[-x^(1/3)-x^(-1/3)]/5
=-[x^(1/3)+x^(-1/3)]/5
=-g(x)
设Y=T(X)=f(x)*g(x)
则有:T(-x)=f(-x)*g(-x)
=[-f(x)]*[-g(x)]
=f(x)g(x)
=T(x)
则y=f(x)*g(x)为偶函数
则有T(X)=f(x)*g(x)图像关于Y轴对称
又y=f(x)*g(x)
={[x^(1/3)-x^(-1/3)]/5}*{[x^(1/3)+x^(-1/3)]/5}
=[x^(2/3)-x^(-2/3)]/5
则任取X1,X2属于(0,+无穷)
且X1>X2
则有:X1^(1/3)>X2^(1/3)>0
0<X1^(-1/3)<X2^(-1/3)
则:f(x1)-f(x2)
=[x1^(2/3)-x1^(-2/3)]/5-{[x2^(2/3)-x2^(-2/3)]/5}
={[x1^(2/3)-x2^(2/3)]-[x1^(-2/3)-x2^(-2/3)]}/5
={[x1^(1/3)+x2^(1/3)]*[x1^(1/3)-x2^(1/3)]-
[x1^(-1/3)+x2^(-1/3)]*[x1^(-1/3)-x2^(-1/3)]}/5<