Question

Find the maximum and minimum values of y = tan $x-2x$.

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=y=\tan x-2x \\ \Rightarrow f\text{'}\left(x\right)={\sec }^{2}x-2 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow {\sec }^{2}x-2=0 \\ \Rightarrow {\sec }^{2}x=2 \\ \Rightarrow \sec x=\pm \sqrt{2} \\ \Rightarrow x=\frac{\mathrm{\pi }}{4}\mathrm{and}\frac{3\mathrm{\pi }}{4} \\ \mathrm{Thus},x=\frac{\mathrm{\pi }}{4}\mathrm{and}x=\frac{3\mathrm{\pi }}{4}\mathrm{are}\mathrm{the}\mathrm{possible}\mathrm{points}\mathrm{of}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima}. \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=2{\sec }^{2}x\tan x \\ \mathrm{At}x=\frac{\mathrm{\pi }}{4}: \\ f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=2{\sec }^2\left(\frac{\mathrm{\pi }}{4}\right)\tan \left(\frac{\mathrm{\pi }}{4}\right)=4>0 \\ \mathrm{So},x=\frac{\mathrm{\pi }}{4}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{The}\mathrm{local}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(\frac{\mathrm{\pi }}{4}\right)=\tan \left(\frac{\mathrm{\pi }}{4}\right)-2\times \frac{\mathrm{\pi }}{4}=1-\frac{\mathrm{\pi }}{2} \\ \mathrm{At}x=\frac{3\mathrm{\pi }}{4}: \\ f\text{'}\text{'}\left(\frac{3\mathrm{\pi }}{4}\right)=2{\sec }^2\left(\frac{3\mathrm{\pi }}{4}\right)\tan \left(\frac{3\mathrm{\pi }}{4}\right)=-4<0 \\ \mathrm{So},x=\frac{3\mathrm{\pi }}{4}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{maximum}. \\ \mathrm{The}\mathrm{local}\mathrm{maximum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(\frac{3\mathrm{\pi }}{4}\right)=\tan \left(\frac{3\mathrm{\pi }}{4}\right)-2\times \frac{3\mathrm{\pi }}{4}=-1-\frac{3\mathrm{\pi }}{2} \end{gathered}$$