Question

設(shè)函數(shù)f(x)=e^x的反函數(shù)為g(x),點(diǎn)P(x₁,y₁),Q(x₂,y₂)分別為函數(shù)f(x)和g(x)圖象上的兩個(gè)動(dòng)點(diǎn).

(1)求函數(shù)h(x)=x²-g(x)的極小值;

(2)設(shè)函數(shù)f(x)的圖象為C₁,g(x)的圖象為C₂,過(guò)點(diǎn)P,Q的直線(xiàn)為l,當(dāng)直線(xiàn)l為曲線(xiàn)C₁和曲線(xiàn)C₂的公切線(xiàn)時(shí),求x₁與x₂滿(mǎn)足的關(guān)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(shù)g(x)=lnx,h(x)=x²-lnx(x＞0).                                                  ?

∵h′(x)=2x-,令h′(x)=0,解得x=±,?

又∵x＞0,∴x=.?

當(dāng)0＜x＜時(shí),h′(x)＜0,∴h(x)在區(qū)間(0,)內(nèi)為減函數(shù);?

當(dāng)x＞時(shí),h′(x)＞0,∴h(x)在區(qū)間(,+∞)內(nèi)為增函數(shù);                                ?

故當(dāng)x=時(shí),函數(shù)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(dāng)直線(xiàn)l為曲線(xiàn)C₁與曲線(xiàn)C₂的公切線(xiàn)時(shí),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(dāng)x₁＞1時(shí), E^x₁＞E,解＞E,可得x₁＜,即1＜x₁＜,?

當(dāng)x₁＜-1時(shí), E^x₁＜,解＜,可得x₁＞,?

即＜x₁＜-1,?

故x₁∈(,-1)∪(1,).