永远的骄傲
幼苗
共回答了13个问题采纳率:84.6% 举报
f(x)=[x^(1/3)-x^(-1/3)]/5,g(X)=[x^(1/3)+x^(-1/3)]/5.
1: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),
∴f(x)是奇函数.
∵x^(-1/3)定义域是x≠0,且在定义域上是减函数,
∴-x^(-1/3)是增函数,而x^(1/3)是R上的增函数,
∴x^(1/3)-x^(-1/3)增函数.
故f(x)有两个增区间(-∞,0)和(0,+∞).
2.f(4)-5f(2)g(2)
=[4^(1/3)-4^(-1/3)]/5-5*{[2^(1/3)-2^(-1/3)]/5}*{[2^(1/3)+2^(-1/3)]/5}
=[4^(1/3)-4^(-1/3)]/5-[4^(1/3)-4^(-1/3)]/5
=0,
同样,有f(9)-5f(3)g(3)=0.
一般地,有f(x^2)-5f(x)g(x)=0,证明如下:
f(x^2)-5f(x)g(x)
=[(x^2)^(1/3)-(x^2)^(-1/3)]/5-5*{[x^(1/3)-x^(-1/3)]/5}*{[x^(1/3)+x^(-1/3)]/5}
=[x^(2/3)-x^(-2/3)]/5-[x^(2/3)-x^(-2/3)]/5
=0.
证毕.
1年前
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