每日导数44

切线问题

已知\(f(x)=\dfrac{2}{x}+\ln x\)的图像在\(x=4\)处的切线为\(y=l(x)\)

(1) 求\(f(x)\)的解析式

(2) 若过点\((a,b)(a<4)\)可作\(f(x)\) 图像的三条切线，证明：\(l(a)<b<f(a)\)

(1) \(f(4)=2\ln 2+\dfrac{1}{2},f^{\prime}(x)=\dfrac{1}{x}-\dfrac{2}{x^2},f^{\prime}(4)=\dfrac{1}{8}\)

则\(y=2\ln 2+\dfrac{1}{2}+\dfrac{1}{8}(x-4)=\dfrac{1}{8}x+2\ln 2\)

(2)设切点为\(\left(x_0,\dfrac{2}{x_0}+\ln x_0\right)\)

则切线为\(y=\left(\dfrac{1}{x_0}-\dfrac{2}{x_0^2}\right)(x-x_0)+\dfrac{2}{x_0}+\ln x_0\)

从而\(b=\left(\dfrac{1}{x_0}-\dfrac{2}{x_0^2}\right)(a-x_0)+\dfrac{2}{x_0}+\ln x_0\)有三个不同的解

考虑\(h(x)=\left(\dfrac{1}{x}-\dfrac{2}{x^2}\right)(a-x)+\dfrac{2}{x}+\ln x-b=a\left(\dfrac{1}{x}-\dfrac{2}{x^2}\right)+\dfrac{4}{x}+\ln x-1-b\)

有三个零点

\(h^{\prime}(x)=\dfrac{(x-4)(x-a)}{x^3}\)

当\(a\leq 0\)时，\(h(x)\)在\((0,4)\)递减，在\((4,+\infty)\)上递增

从而\(h(x)\)至多有两个零点，不题

当\(0<a<4\)时，\(h(x)\)在\((0,a)\)和\((4,+\infty)\)上递增，在\((a,4)\)上递减

因\(h(x)\)有三个不同零点，则一定有

\(h(a)=\dfrac{2}{a}+\ln a-b>0\)，即\(f(a)>b\)

\(h(4)=\dfrac{a}{8}+2\ln 2-b<0\)

即\(b>\dfrac{a}{8}+2\ln 2=l(a)\)