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

Question

Christopher Allers · Accepted Answer

Increasing.

#d/dxf=((sqrt3+1)e^(pi/6)-pi)/(2(pi-e^(pi/6))^2)#. Numerator at x = #pi/6# is 1.470.....>0 The first derivative is > 0..

Ezekiel Alexander · Answer

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

The derivative of $ f(x) $ with respect to $ x $ is given by:

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

Evaluating $ f'(x) $ at $ x = \frac{\pi}{6} $, we have:

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

$$ = \frac{-\frac{1}{2}\left(\pi - e^{\frac{\pi}{6}}\right) - \frac{\sqrt{3}}{2}\left(-e^{\frac{\pi}{6}}\right)}{\left(\pi - e^{\frac{\pi}{6}}\right)^2} $$

$$ = \frac{-\frac{1}{2}\pi + \frac{1}{2}e^{\frac{\pi}{6}} + \frac{\sqrt{3}}{2}e^{\frac{\pi}{6}}}{\left(\pi - e^{\frac{\pi}{6}}\right)^2} $$

Now, evaluate the sign of $ f'\left(\frac{\pi}{6}\right) $ to determine if $ f(x) $ is increasing or decreasing at $ x = \frac{\pi}{6} $. If $ f'\left(\frac{\pi}{6}\right) > 0 $, then $ f(x) $ is increasing at $ x = \frac{\pi}{6} $. If $ f'\left(\frac{\pi}{6}\right) < 0 $, then $ f(x) $ is decreasing at $ x = \frac{\pi}{6} $.