数学分析味道很浓的一道题,可以当作找点问题的典型.
已知函数\(f(x)=e^x-ax^2-\cos x-\ln(x+1)\)
\((1)\) 若\(a=1\),求证:\(f(x)\)的图像与\(x\)轴相切与原点
\((2)\) 若函数\(f(x)\)在区间\((-1,0),(0,+\infty)\)上各恰有一个极值点,求实数\(a\)的取值范围.
解
\((1)\) \(a=1,f(x)=e^x-x^2-\cos x-\ln(x+1)\)
\(f^{\prime}(x)=e^x-2x+\sin x-\dfrac{1}{x+1},f^{\prime}(0)=1-1=0\)
而\(f(0)=0\),从而在原点的切线跟\(x\)轴相切
\((2)\) \(f^{\prime}(x)=e^x-2ax+\sin x-\dfrac{1}{x+1},f^{\prime\prime}(x)=e^x+\cos x-2a+\dfrac{1}{(x+1)^2}\)
分析:先考虑\(x\in(-1,0)\)上,\(\lim\limits_{x\to -1^{+}}f^{\prime}(x)=-\infty,\lim\limits_{x\to 0}f^{\prime}(x)=0,\lim\limits_{x\to-1^{+}}f^{\prime\prime}(x)\to +\infty,\lim\limits_{x\to 0}f^{\prime\prime}(x)=3-2a\)
因\(\lim\limits_{x\to-1^{+}}f^{\prime}(x)\to-\infty,\lim\limits_{x\to 0}f^{\prime}(x)= 0\)分析得知,\(f^{\prime}(x)\)在\((-1,0)\)上一定是先增再减的
并且最大值要大于\(0\),但是因为\(\lim\limits_{x\to -1^{+}}f^{\prime}(x)=-\infty,\lim\limits_{x\to 0}f^{\prime}(x)=0\),这个条件如果说明了
先增再减是无需说明最大值大于\(0\)的,这是显而易见的.
从而也就是说明\(f^{\prime\prime}(x)\)先是正的再是负的,从而也说明了还得再次求导研究其正负
进而分析出\(\lim\limits_{x\to 0}f^{\prime\prime}(x)< 0\),即\(3-2a< 0\)得\(a> \dfrac{3}{2}\).
在\(a>\dfrac{3}{2}\)的情形下,\(\lim\limits_{x\to +\infty}f^{\prime}(x)=+\infty,f^{\prime\prime}(x)\to +\infty\)而\(f^{\prime}(0)=0,f^{\prime\prime}(0)=3-2a<0\)
那么\(f^{\prime}(x)\)只有先减,再增才能使得其在\((0,+\infty)\)上只有一个零点.
从而\(f^{\prime\prime}(x)\)一定先小于\(0\)再大于\(0\),而\(f^{\prime\prime}(0)=3-2a<0\)
则\(f^{\prime\prime}(x)\)要么先减再增,要么单调递增,如果还有其他的情况那就太复杂了不适合高中生.
而\(f^{\prime\prime\prime}(x)=e^x-\sin x-\dfrac{2}{(x+1)^3}\),\(f^{\prime\prime\prime}(0)=-1<0,\lim\limits_{x\to+\infty}f^{\prime\prime\prime}(x)\to+\infty\)
因此要考虑下\(f^{(4)}(x)\)的情况
下面开始书写过程:
Case1.当\(a\leq \dfrac{3}{2}\)时,\(f^{\prime\prime\prime}(x)=e^x-\sin x-\dfrac{2}{(x+1)^3}<0\)
从而\(f^{\prime\prime\prime}(x)\)在\((-1,0)\)上为负,即\(f^{\prime\prime}(x)\)单调递减,即\(f^{\prime\prime}(x)>3-2a>0\)
从而\(f^{\prime}(x)\)单调递增,在\((-1,0)\)没有极值点,不合题.
Case2.当\(a>\dfrac{3}{2}\)时,则一定存在\(x_0\in(-1,0)\)使得\(f^{\prime\prime}(x_0)=0\)
从而\(f^{\prime\prime}(x)\)在\((-1,x_0)\)大于\(0\),在\((x_0,0)\)上小于\(0\)
从而\(f^{\prime}(x)\)在\((-1,x_0)\)上单调递增,在\((x_0,0)\)上单调递减
而:\(\lim\limits_{x\to -1^{+}}f^{\prime}(x)=-\infty,\)\(\lim\limits_{x\to 0}f^{\prime}(x)=0\)
则说明\(f^{\prime}(x)\)在\((-1,0)\)上一定存在唯一的零点
则\(f(x)\)在\((-1,0)\)上有唯一极值点
再看\(x\in(0,+\infty)\)
因\(f^{\prime\prime\prime}(x)=e^x-\sin x-\dfrac{2}{(x+1)^3}\)
\(f^{\prime\prime\prime}(0)=-1<0,\lim\limits_{x\to+\infty}f^{\prime\prime\prime}(x)\to+\infty\)
\(f^{(4)}(x)=e^x-\cos x+\dfrac{6}{(x+1)^4}>0\),则\(f^{\prime\prime\prime}(x)\)单调递增
因\(f^{\prime\prime\prime}(0)=-1<0,\lim\limits_{x\to+\infty}f^{\prime\prime\prime}(x)\to+\infty\),且是单调递增的
则一定有\(x_{1}\)使得\(f^{\prime\prime\prime}(x_1)=0\)
从而\(f^{\prime\prime}(x)\)在\((0,x_1)\)递减,在\((x_1,+\infty)\)上递增
从而\(f^{\prime}(x)\)在\((0,+\infty)\)上一定只有一个零点.
综上:\(a>\dfrac{3}{2}\).