隐藏的极值点偏移
已知函数\(f(x)=\dfrac{1}{2}x^2-x-a\ln(x+1)\)
\((1)\)讨论函数\(f(x)\)的单调性
\((2)\)当\(a>0\)时,若\(m\)为函数的正零点,证明:\(m>2\sqrt{a+1}\)
解
\((1)\)由题得\(x>-1\) \(f^{\prime}(x)=x-\dfrac{a}{x+1}-1=\dfrac{x^2-(a+1)}{x+1}\)
若\((a+1)\leq 1\),则\(f^{\prime}(x)\geq 0\),则\(f(x)\)在定义域上单调递增
若\(a>0,f^{\prime}(x)=0\)得\(x_1=\sqrt{a+1}>1,x_2=-\sqrt{a+1}<-1\)
从而不难知道\(f(x)\)在\((-1,\sqrt{a+1})\)上单调递减,\((\sqrt{a+1},+\infty)\)上单调递增
综上:\(a\leq 0\)时,\(f(x)\)在第定义域上单调递增, \(a>0\)时,\(f(x)\)在\((-1,\sqrt{a+1})\)上单调递
减, \((\sqrt{a+1},+\infty)\) 上单调递增
\((2)\) 不难发现\(f(0)=0\),而\(f(x)\)在\((0,
\sqrt{a+1})\)上单调递减,\((\sqrt{a+1},+\infty)\) 上单调递增.\(x\to +\infty,f(x)\to +\infty\)
从而一定会存在一个正根.并且此正根比\(\sqrt{a+1}\)大.
则要证\(m>2\sqrt{{a+1}}>2\)
即证:\(\dfrac{m^2}{4}-1>a\)
由题\(f(m)=\dfrac{1}{2}m^2-m-a\ln(m+1)=0\)
即证:\(\dfrac{1}{2}m^2-m=a\ln(m+1)<\left(\dfrac{m^2-4}{4}\right)\ln(m+1)\)
即证:\(2m(m-2)<(m-2)(m+2)\ln(m+1)\)
又因\(m>2\)
即证:\(2m<(m+2)\ln(m+1)\)
即证:\(\dfrac{2m}{m+2}-\ln(m+1)<0\)
记\(g(m)=\dfrac{2m}{m+2}-\ln(m+1)\)
\(g^{\prime}(m)=\dfrac{-m^2}{(m+2)^2(m+1)}<0\)
从而\(g(m)<g(2)=1-\ln 3<0\)
得证.