函数f(x)=7x3+2x+1,则不等式f(x)+f(x-1)>2的解集 .
【答案】
分析:设F(x)=f(x)+f(x-1),利用求导法则求出F(x)的导函数,根据导函数恒大于0,得到函数F(x)为增函数,再由x=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/0.png)
时,f(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/1.png)
)+f(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/2.png)
-1)=2,利用增减性即可得出所求不等式的解集.
解答:解:设F(x)=f(x)+f(x-1),
由f′(x)=21x
2+2>0,f′(x-1)=21(x-1)
2+2>0,
得到F′(x)>0,即F(x)为增函数,
又当x=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/3.png)
时,F(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/4.png)
)=f(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/5.png)
)+f(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/6.png)
-1)=7×(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/7.png)
)
3+2×
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/8.png)
+1+7×(-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/9.png)
)
3+2×(-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/10.png)
)+1=2,
则不等式f(x)+f(x-1)>2的解集为(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/11.png)
,+∞).
故答案为:(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025123532125254469/SYS201310251235321252544013_DA/12.png)
,+∞)
点评:此题考查了其他不等式的解法,解决此类问题的关键是正确利用函数的单调性,结合不等式的解法解出x的范围,此知识点是高考考查的重点之一.