Question

设函数f(x)=e^x的反函数为g(x),点P(x₁,y₁),Q(x₂,y₂)分别为函数f(x)和g(x)图象上的两个动点.

(1)求函数h(x)=x²-g(x)的极小值;

(2)设函数f(x)的图象为C₁,g(x)的图象为C₂,过点P,Q的直线为l,当直线l为曲线C₁和曲线C₂的公切线时,求x₁与x₂满足的关系式及x₁的取值范围.

Answer 1

解:(1)∵f(x)=E^x,即y=E^xx=lny,?

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

∵h′(x)=2x-,令h′(x)=0,解得x=±,?

又∵x＞0,∴x=.?

当0＜x＜时,h′(x)＜0,∴h(x)在区间(0,)内为减函数;?

当x＞时,h′(x)＞0,∴h(x)在区间(,+∞)内为增函数;                                ?

故当x=时,函数h(x)取得极小值,且极小值为?

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

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

∴当直线l为曲线C₁与曲线C₂的公切线时,它的方程为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.?

当x₁＞1时, E^x₁＞E,解＞E,可得x₁＜,即1＜x₁＜,?

当x₁＜-1时, E^x₁＜,解＜,可得x₁＞,?

即＜x₁＜-1,?

故x₁∈(,-1)∪(1,).