Is #f(x)=e^-xcos(-x)-sinx/(pi-e^x)# increasing or decreasing at #x=pi/6#?

Question

Charles Anctil · Accepted Answer

#f(x)# is decreasing at #x=\pi/6#

Given function:

#f(x)=e^{-x}\cos(-x)-\frac{\sin x}{\pi-e^x}#

#f(x)=e^{-x}\cosx+\frac{\sin x}{e^x-\pi}#

Differentiating above function w.r.t. #x# using chain rule & quotient rule as follows

#f'(x)=d/dx(e^{-x}\cosx+\frac{\sin x}{e^x-\pi})#

#=e^{-x}(-\sinx)+\cos x(-e^{-x})+\frac{(e^x-\pi)\cos x-\sin x(e^x)}{(e^x-\pi)^2}#

#=-e^{-x}(\sinx+\cos x)+\frac{e^x(\cos x-\sin x)-\pi\cos x}{(e^x-\pi)^2}#

Setting #x=\pi/6# in above expression we get

#f'(\pi/6)#

#=-e^{-\pi/6}(\sin(\pi/6)+\cos (\pi/6))+\frac{e^{\pi/6}(\cos (\pi/6)-\sin (\pi/6))-\pi\cos (\pi/6)}{(e^{\pi/6}-\pi)^2}#

#=-0.80921-0.9953#

#=-1.8045#

Since #f'(\pi/6)<0# hence the given function #f(x)# is decreasing at #x=\pi/6#

Emily Ammerman · Answer

undefined To determine whether $ f(x) = e^{-x}\cos(-x) - \frac{\sin x}{\pi - e^x} $ is increasing or decreasing at $ x = \frac{\pi}{6} $, we need to find the sign of its derivative at that point.

Calculate the derivative of $ f(x) $ with respect to $ x $:

$$ f'(x) = -e^{-x}\cos(-x) + e^{-x}\sin(-x) - \frac{-e^x\sin x - e^x\cos x}{(\pi - e^x)^2} $$

Evaluate $ f'(x) $ at $ x = \frac{\pi}{6} $:

$$ f'\left(\frac{\pi}{6}\right) = -e^{-\frac{\pi}{6}}\cos\left(-\frac{\pi}{6}\right) + e^{-\frac{\pi}{6}}\sin\left(-\frac{\pi}{6}\right) - \frac{-e^{\frac{\pi}{6}}\sin\frac{\pi}{6} - e^{\frac{\pi}{6}}\cos\frac{\pi}{6}}{(\pi - e^{\frac{\pi}{6}})^2} $$

$$ = -e^{-\frac{\pi}{6}}\cos\left(\frac{\pi}{6}\right) + e^{-\frac{\pi}{6}}\sin\left(\frac{\pi}{6}\right) - \frac{-e^{\frac{\pi}{6}}\sin\frac{\pi}{6} - e^{\frac{\pi}{6}}\cos\frac{\pi}{6}}{\left(\pi - e^{\frac{\pi}{6}}\right)^2} $$

$$ \approx -0.878 $$

Since $ f'\left(\frac{\pi}{6}\right) $ is negative, $ f(x) $ is decreasing at $ x = \frac{\pi}{6} $.