同构、隐零点
已知函数\(f(x)=x\ln x+(t-1)x-t\)
(1)当\(t=0\),讨论\(f(x)\)的极值
(2)若\(F(x)=f(x)-\dfrac{e^x}{e^t}\)有两个不同的极值点,求\(t\)的取值范围
解
(1)\(f(x)=x\ln x-x\),\(f^{\prime}(x)=\ln x\),则得到\(f(x)\)有唯一的极值点,并且是\(x=1\)极小值点
,并且\(f(1)=-1\)
(2)
法一:隐零点处理
\(F(x)=x\ln x+(t-1)x-t-e^{x-t}\),\(F^{\prime}(x)=\ln x+t-e^{x-t}\)
记\(\varphi(x)=\ln x+t-e^{x-t}\),\(\varphi^{\prime}(x)=\dfrac{1}{x}-e^{x-t}\)单调递增
\(x\to 0,\varphi(x)\to-\infty,x\to+\infty,\varphi(x)\to-\infty\),则有唯一的零点\(x_0\),使得\(\varphi^{\prime}(x_0)=0\)
并且\(x=x_0\)是\(\varphi(x)\)的极大值点
则要使得\(\varphi(x)\)有两个零点,要有\(\varphi(x_0)>0\)
\(\dfrac{1}{x_0}=e^{x_0-t}\),即\(t=x_0+\ln x_0\)
则\(\varphi(x_0)=\ln x_0+t-e^{x_0-t}=\ln x_0+x_0+\ln x_0-\dfrac{1}{x_0}=2\ln x_0+x_0-\dfrac{1}{x_0}>0\)
因\(2\ln x_0+x_0-\dfrac{1}{x_0}\)单调递增
则\(x_0>1\)
从而\(t=x_0+\ln x_0>1\)
法二:同构处理
\(F^{\prime}(x)=0\)有\(\ln x+t-e ^{x-t}=0\)
即\(\ln x=e^{x-t}-t\)
即\(\ln x+x=e^{x-t}+x-t\)
即\(e^{\ln x}+\ln x=e^{x-t}+x-t\)
考虑\(\gamma(x)=e^{x}+x\)
即\(\gamma(\ln x)=\gamma(x-t)\)
因\(\gamma(x)\)单调递增
则\(\ln x=x-t\)
即\(y=\ln x\)与\(y=x-t\)有两个交点
而\(y=\ln x\)与\(y=x-1\)相切
则要使得有两个交点
要有\(t>1\)
标签:prime,导数,ln,dfrac,每日,varphi,零点,83,gamma From: https://www.cnblogs.com/manxinwu/p/18069739