Is #f(x)=sinx-cos(pi-x)# increasing or decreasing at #x=pi/3#?

Question

Evelyn Agard · Accepted Answer

Decreasing.

Start by expanding #cos(pi - x)#. #f(x) = sinx- cospicosx + sinpisinx# #f(x) = sinx - (-1cosx) + 0(sinx)# #f(x) = sinx + cosx# Distinguish: #f'(x) = cosx - sinx# Consider your point in relation to the derivative. #f(pi/3) = cos(pi/3) - sin(pi/3) = 1/2 - sqrt(3)/2 = (1 - sqrt(3))/2# This is negative, so the function is decreasing at #x= pi/3#. I hope this is useful!

Ezekiel Alexander · Answer

undefined To determine whether $ f(x) = \sin(x) - \cos(\pi - x) $ is increasing or decreasing at $ x = \frac{\pi}{3} $, we need to analyze its derivative at that point.

First, find the derivative of $ f(x) $:
$$ f'(x) = \cos(x) + \sin(\pi - x) $$

Now, evaluate the derivative at $ x = \frac{\pi}{3} $:
$$ f'\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + \sin\left(\pi - \frac{\pi}{3}\right) $$
$$ = \cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{2\pi}{3}\right) $$
$$ = \frac{1}{2} + \frac{\sqrt{3}}{2} $$

Since $ \frac{1}{2} + \frac{\sqrt{3}}{2} > 0 $, $ f'(x) $ is positive at $ x = \frac{\pi}{3} $.

Therefore, $ f(x) $ is increasing at $ x = \frac{\pi}{3} $.