Question

設函數f(x)=e^x的反函數為g(x),點P(x₁,y₁),Q(x₂,y₂)分別為函數f(x)和g(x)圖象上的兩個動點.

(1)求函數h(x)=x²-g(x)的極小值;

(2)設函數f(x)的圖象為C₁,g(x)的圖象為C₂,過點P,Q的直線為l,當直線l為曲線C₁和曲線C₂的公切線時,求x₁與x₂滿足的關系式及x₁的取值范圍.

Answer 1

解:(1)∵f(x)=E^x,即y=E^xx=lny,?

∴f(x)的反函數g(x)=lnx,h(x)=x²-lnx(x＞0).                                                  ?

∵h′(x)=2x-,令h′(x)=0,解得x=±,?

又∵x＞0,∴x=.?

當0＜x＜時,h′(x)＜0,∴h(x)在區間(0,)內為減函數;?

當x＞時,h′(x)＞0,∴h(x)在區間(,+∞)內為增函數;                                ?

故當x=時,函數h(x)取得極小值,且極小值為?

h()=()²-ln=-ln.                                                                   ?

(2)∵f′(x)=E^x,g′(x)=,又P(x₁,E^x₁),Q(x₂,lnx₂),

∴當直線l為曲線C₁與曲線C₂的公切線時,它的方程為y-E^x₁=E^x₁ (x-x₁)                    ①?

或y-lnx₂=(x-x₂),                                                                                     ②?

由①得,y=E^x₁·x+E^x₁(1-x₁),由②得,y=·x+lnx₂-1,

∴= E^x₁x₂= E^-x₁,即x₂=E^-x₁.                                                                             ?

∴E^x₁ (1-x₁)=lnx₂-1=lnE^-x₁-1E^x₁(1-x₁)=-x₁-1E^x₁=.?

又∵E^x₁＞0,∴＞0x₁＜-1或x₁＞1.?

當x₁＞1時, E^x₁＞E,解＞E,可得x₁＜,即1＜x₁＜,?

當x₁＜-1時, E^x₁＜,解＜,可得x₁＞,?

即＜x₁＜-1,?

故x₁∈(,-1)∪(1,).