Find the value of $ \displaystyle x$ between $ \displaystyle 0$ and $ \displaystyle \frac{\pi}{2}$ for which the curve $\displaystyle y={{e}^{{\sqrt{3}x}}}\cos x$ has a stationary point. Determine whether it is a maximum or minimum point.
Solution
$ \displaystyle y={{e}^{{\sqrt{3}x}}}\cos x$
$ \displaystyle \frac{{dy}}{{dx}}=-\sin x\cdot {{e}^{{\sqrt{3}x}}}+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}\cos x$
$ \displaystyle \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$
$\displaystyle \frac{{dy}}{{dx}}=0$, when
$ \displaystyle {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)=0$
Since $ \displaystyle {{e}^{{\sqrt{3}x}}}>0$ for every $ \displaystyle x\in R$,
$ \displaystyle \sqrt{3}\cos x-\sin x=0$
$ \displaystyle \therefore \sqrt{3}\cos x=\sin x$
$ \displaystyle \therefore \tan x=\sqrt{3}\Rightarrow x=\frac{\pi }{3}$
$ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x)+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$
$ \displaystyle \ \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x+3\cos x-\sqrt{3}\sin x)$
$ \displaystyle \ \ \ \ \ \ =2{{e}^{{\sqrt{3}x}}}(\cos x-\sqrt{3}\sin x)$
$ \displaystyle {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{3}}}}=2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}(\cos \frac{\pi }{3}-\sqrt{3}\sin \frac{\pi }{3})$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}\left( {\frac{1}{2}-\frac{3}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =-2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}$
$ \displaystyle \therefore {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{3}}}}$ $ \displaystyle <$ $ \displaystyle 0$
Therefore the stationary point is a maximum turning point.
$ \displaystyle y={{e}^{{\sqrt{3}x}}}\cos x$
$ \displaystyle \frac{{dy}}{{dx}}=-\sin x\cdot {{e}^{{\sqrt{3}x}}}+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}\cos x$
$ \displaystyle \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$
$\displaystyle \frac{{dy}}{{dx}}=0$, when
$ \displaystyle {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)=0$
Since $ \displaystyle {{e}^{{\sqrt{3}x}}}>0$ for every $ \displaystyle x\in R$,
$ \displaystyle \sqrt{3}\cos x-\sin x=0$
$ \displaystyle \therefore \sqrt{3}\cos x=\sin x$
$ \displaystyle \therefore \tan x=\sqrt{3}\Rightarrow x=\frac{\pi }{3}$
$ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x)+\sqrt{3}\cdot {{e}^{{\sqrt{3}x}}}(\sqrt{3}\cos x-\sin x)$
$ \displaystyle \ \ \ \ \ \ ={{e}^{{\sqrt{3}x}}}(-\sqrt{3}\sin x-\cos x+3\cos x-\sqrt{3}\sin x)$
$ \displaystyle \ \ \ \ \ \ =2{{e}^{{\sqrt{3}x}}}(\cos x-\sqrt{3}\sin x)$
$ \displaystyle {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{3}}}}=2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}(\cos \frac{\pi }{3}-\sqrt{3}\sin \frac{\pi }{3})$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}\left( {\frac{1}{2}-\frac{3}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =-2{{e}^{{\frac{{\sqrt{3}\pi }}{3}}}}$
$ \displaystyle \therefore {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{3}}}}$ $ \displaystyle <$ $ \displaystyle 0$
Therefore the stationary point is a maximum turning point.
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