Question

求解高一数学题，过程

1.已知f(x)=3|x|,求f(x)的单调区间
2.已知已知f(x)=x^3+x，x∈R，判断f(x)的单调性并证明。

Answer 1

1、当x>=0时，f(x)=3x,单调增；
当x<=0时，f(x)=-3x,单调减；
2、单调增，设x1>x2
f(x1)-f(x2)=(x1)³-(x2)³+(x1-x2)
=(x1-x2)(x1²+x2²+x1*x2)+(x1-x2)
=(x1-x2)(x1²+x2²+x1*x2+1)
=(x1-x2)[(x1+0.5x2）²+0.75x2²+1]
因为x1>x2,(x1+0.5x2）²+0.75x2²+1恒大于0
所以f(x1)-f(x2)>0
即该函数单调增。

Answer 2

1.当x>0时，f(x)=3x,
在（0，+∞）上是单调递增函数
当x<0时，f(x)=-3x
在（-∞，0）上是单调递减函数
当x=0时，f(x)=0,是常数

2. 方法一：
f'(x)=3*x^2+1>0，故单调增；
方法二：
在R上去x1,x2,且x1<x2
f(x1)-f(x2)=x1^3+x1-(x2^3+x2)
=(x1-x2)(x1^2+x1x2+x2^2)+(x1-x2)
=(x1-x2)[(x1+1/2 x2)^2+3/4 x2^2+1]
因为 x1-x2<0,(x1+1/2 x2)^2+(3/4)*x2^2+1>0
所以 f(x1)-f(x2)<0
即 f(x1)<f(x2)
所以f(x))=x^3+x在R上单调递增

Answer 3

(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
又X1^(1/3)>X2^(1/3)>0
0<X1^(-1/3)<X2^(-1/3)
则有: x1^(1/3)+x2^(1/3)>0
x1^(1/3)-x2^(1/3)>0
x1^(-1/3)+x2^(-1/3)>0
x1^(-1/3)-x2^(-1/3)<0
则:[x1^(1/3)+x2^(1/3)]*[x1^(1/3)-x2^(1/3)]>0,
[x1^(-1/3)+x2^(-1/3)]*[x1^(-1/3)-x2^(-1/3)]<0
则f(x1)-f(x2)
={[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>0
即:X1>X2>0时,f(x1)>f(x2)
则T(X)在(0,+无穷)上单调递增
又y=T(X)=f(x)*g(x)为偶函数
则有T(X)在(-无穷,0)上单调递减

(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-{[2^(1/3)-2^(-1/3)]*[2^(1/3)+2^(-1/3)]/5}
=[4^(1/3)-4^(-1/3)]/5-[4^(1/3)-4^(-1/3)]/5
=0
同理可得:f(9)-5f(3)*g(3)=0
归纳得：f(x^2)－5f(x)g(x)＝0

Answer 4

1，讨论：

当x>0时，f（x）=3x,f(x)递增
当x<0,f(x)=-3x,f(x)递减

因此增区间为（0，+∞）；
减区间为（-∞，0]

2，

f'(x)=3x^2+1>0恒成立
因此f(x)在R上递增

Answer 5

1.画图可知（-∞，0）单调递减 （0，+∞）单调递增
2。求导数 f′（x）=3x^2+1 可知f′（x）恒大于0
所以单调区间为全体实数 单调递增