Question

已知函數(shù)f(x)=x+1,設(shè)g₁(x)=f(x),g_n(x)=f(g_n-1(x))(n＞1,n∈N^*).

(1)求g₂(x),g₃(x)的表達(dá)式,并猜想g_n(x)(n∈N^*)的表達(dá)式(直接寫出猜想結(jié)果);

(2)若關(guān)于x的函數(shù)y=x²+(x)(n∈N^*)在區(qū)間(-∞,-1]上的最小值為6,求n的值.(符號“”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當(dāng)-≥-1,即n≤2時,函數(shù)y=(x+)²+在區(qū)間(-∞,-1]上是減函數(shù).

∴當(dāng)x=-1時,y_min==6,即n²-n-10=0,該方程無整數(shù)解.                   

②當(dāng)-＜-1,即n＞2時,y_min==6,解得n=4.                               

綜上,n=4.