Is #f(x)= cos(x+(5pi)/4) # increasing or decreasing at #x=-pi/4 #?

Question

Is #f(x)= cos(x+(5pi)/4)   # increasing or decreasing at #x=-pi/4   #?

Christopher Agresta · Accepted Answer

See explanation.

Generally if the function #f(x)# has the derrivative #f^'(x_0)# then we can say that: #f(x)# is increasing at #x_0# if #f^'(x_0)>0# #f(x)# is decreasing at #x_0# if #f^'(x_0)<0# #f^'(x)# may have an extremum at #x_0# if #f^'(x_0)=0# (additional test is required) In the example provided, we have: #f^'(x)=-sin(x+(5pi)/4)# #f^'(x_0)=-sin(-pi/4+(5pi)/4)=-sin(pi)=0# #f^'(x_0)=0#, so #f(x)# has either an extremum, or an inflection point. To check if the point is extremum we have to check if the first derivative changes sign at #x_0#. graph{

Elijah Abram · Answer

undefined To determine whether $ f(x) = \cos\left(x + \frac{5\pi}{4}\right) $ is increasing or decreasing at $ x = -\frac{\pi}{4} $, we need to examine the derivative of the function at that point.

The derivative of $ \cos(x) $ is $ -\sin(x) $. Thus, the derivative of $ f(x) $ with respect to $ x $ is $ -\sin\left(x + \frac{5\pi}{4}\right) $.

At $ x = -\frac{\pi}{4} $, we need to evaluate $ -\sin\left(-\frac{\pi}{4} + \frac{5\pi}{4}\right) $.

$ -\frac{\pi}{4} + \frac{5\pi}{4} = \pi $

$ \sin(\pi) = 0 $

Since the sine function equals $ 0 $ at $ x = \pi $, the derivative of $ f(x) $ is $ 0 $ at $ x = -\frac{\pi}{4} $.

Therefore, the function $ f(x) $ is neither increasing nor decreasing at $ x = -\frac{\pi}{4} $.