求最值

点 (0,1) 到直线 y=k(x+1) 的距离最大值。

$$d=\frac{|k-1|}{\sqrt{k^2+1}}$$ $$\begin{array}{r}设k=\tan \alpha \\ \frac{|\frac{\sin \alpha }{\cos \alpha }-1|}{|\frac{1}{\cos \alpha }}=|\frac{\frac{\sin \alpha }{\cos \alpha }-1}{\frac{1}{\cos \alpha }}|=|\sin \alpha -\cos \alpha |\le \sqrt{2}\end{array}$$

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$\sqrt{1-x^2}$ 设 $x=\sin \alpha ,\sqrt{1-{\sin }^2\alpha }=|\cos \alpha |$

\\设 x = \tan \alpha \\&\sqrt{1+\tan^2\alpha}=\frac1{|\cos\alpha|}\end{aligned}$$ ---

\begin{aligned}
f(x)=\frac{\sqrt{16k^2}}{|3k+1|}\
设4k=\tan \alpha\
f(x)=\frac{\sqrt{\tan^2\alpha+1}}{|\frac34\tan\alpha+1|}=\frac{\frac1{|\cos\alpha|}}{|\frac{3\sin\alpha}{4\cos\alpha}+1|}=\left|\frac{\frac1{\cos\alpha}}{\frac{3\sin\alpha}{4\cos\alpha}+1}\right|\
f(x)=\left|\frac4{3\sin\alpha+4\cos\alpha}\right|\geq\frac{4}{\sqrt{3^{2}+4^{2}}}
\end{aligned}