例说导数零点问题的处理策略

JIETIJIQIAOYUFANGFA1152015.4◎315100【】、．【】．、．12013fx=ax－52+6lnxa∈Ｒy=fx1f1y06．1a2fx．1a=12．2a=12fx=12x－52+6lnxf'x=x－5+6x=x－2x－3x．f'x=0x1=2x2=3．0＜x＜2x＞3f'x＞0fx023+∞2＜x＜3f'x＜0fx23．fxx=2f2=92+6ln2x=3f3=2+6ln3．、lnxex．22013lCy=lnxx10．1l210Cl．1ly=x－1．2gx=x－1－lnxxx＞0x≠1gx＞0．g'x=x2－1+lnxx2x=1g'1=0．0＜x＜1x2－1＜0lnx＜0g'x＜0gxx＞1x2－1＞0lnx＞0g'x＞0gx．x＞0x≠1gx＞g1=0Cl．、“”．3fx=lnx+a+x2fxafxlne2．fx－a+∞f'x=1x+a+2x=2x2+2ax+1x+a．gx=2x2+2ax+1fxgx＜0－a+∞－a2≤－ag－a{＜0－a2＞－ag－a2{＜0a＞槡2．a＞槡2g－a=12x2+2ax+1=0－a+∞x1x2－a＜x1＜x2．－a＜x＜x1x＞x2gx＞0f'x＞0fx－ax1x2+∞x1＜x＜x2gx＜0f'x＜0fxx1x2．fxx=x1x=x2．x1x22x2+2ax+1=0x1+x2=－ax1·x2=12fxfx1+fx2=lnx1+a+x21+lnx2+a+x22=lnx1x2+ax1+x2+a2+x1+x22－2x1x2=ln12+a2－1＞ln12+1=lne2．、．42010afx=ex－2x+2ax∈Ｒ．12a＞ln2－1x＞0ex＞x2－2ax+1．gx=ex－x2+2ax－1a＞ln2－1x＞0gx＞0．g'x=ex－2x+2ahx=ex－2x+2ah'x=ex－2h'x=0x=ln2．x＜ln2h'x＜0hxx＞ln2h'x＞0hx．hx＞hln2=2－2ln2+2a＞0g'x＞0y=gxＲa＞ln2－1x＞0gx＞g0=0ex＞x2－2ax+1．、“”．．．