Question

已知函数f(x)=x+1,设g₁(x)=f(x),g_n(x)=f(g_n-1(x))(n＞1,n∈N^*).

(1)求g₂(x),g₃(x)的表达式,并猜想g_n(x)(n∈N^*)的表达式(直接写出猜想结果);

(2)若关于x的函数y=x²+(x)(n∈N^*)在区间(-∞,-1]上的最小值为6,求n的值.(符号“”表示求和,例如:=1+2+3+…+n).

Answer 1

解:(1)∵g₁(x)=f(x)=x+1,

∴g₂(x)=f［g₁(x)］=f(x+1)=(x+1)+1=x+2,                                         

g₃(x)=f［g₂(x)］=f(x+2)=(x+2)+1=x+3.                                          

∴猜想g_n(x)=x+n.                                                           

(2)∵g_n(x)=x+n,

∴(x)=g₁(x)+g₂(x)+…+g_n(x)=nx+.                                 

∴y=x²+(x)=x²+nx+                                            

=(x+)²+.                                                        

①当-≥-1,即n≤2时,函数y=(x+)²+在区间(-∞,-1]上是减函数.

∴当x=-1时,y_min==6,即n²-n-10=0,该方程无整数解.                   

②当-＜-1,即n＞2时,y_min==6,解得n=4.                               

综上,n=4.