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

Question

Charles Arocha · Accepted Answer

Locally decreasing function.

#f'(x) = e^x - sin x - e^x sin x - e^x cos x# #f'(pi/6) = e^(pi/6)(1 - sin frac{pi}{6} - cos frac{pi}{6}) - sin frac{pi}{6} # #f'(pi/6) = e^(pi/6)(1 - 1/2 - sqrt 3 / 2) - 1/2 < 0 # #e^(pi/6)(1/2 - sqrt 3 / 2) < 1/2# #e^(pi/6)(1 - sqrt 3) < 1# #e^(pi/6)(-1) < 1/(sqrt 3 - 1)# #e^(pi/6) > -1/(sqrt 3 - 1)# favorable > unfavorable

Ezekiel Alexander · Answer

undefined To determine if $ f(x) = e^x + \cos(x) - e^x\sin(x) $ is increasing or decreasing at $ x = \frac{\pi}{6} $, we need to evaluate the first derivative of the function at that point. If the derivative is positive, the function is increasing; if negative, it's decreasing.

$ f'(x) = e^x - \sin(x) - e^x\sin(x) + e^x\cos(x) $

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

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

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

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

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

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

Since $ f'\left(\frac{\pi}{6}\right) > 0 $, the function $ f(x) $ is increasing at $ x = \frac{\pi}{6} $.