首先求定義域得{x| x>0},再對(duì)函數(shù)求導(dǎo)： f'(x)=p+p/x^2-2/x=(px^2-2x+p)/x^2
因?yàn)楹瘮?shù)在定義域內(nèi)為單調(diào)函數(shù),所以f'(x)要么大于等于0,要么小于等于0
若f'(x)大于等于0,則由x>0知p必不為0（若為0則,-2x>=0,與定義域矛盾）于是有,
函數(shù)y=px^2-2x+p的判別式4-4p^2<=0解得p>=1或p<=-1
若f'(x)小于等于0,則當(dāng)p=0時(shí), f'(x)<=0成立所以p可以為0,
當(dāng)p不為0時(shí),要使 f'(x)<=0,則必須使p<0,且判別式4-4p^2<=0解得p<=-1
綜上便可得p的范圍：當(dāng)函數(shù)f(x)為增函數(shù)時(shí),p>=1或p<=-1；當(dāng)函數(shù)f(x）為減函數(shù)時(shí),p<=-1
或p=0