Is #f(x)= 4xcos(3x-(5pi)/4) # increasing or decreasing at #x=-pi/4 #?

Question

Is #f(x)= 4xcos(3x-(5pi)/4)   # increasing or decreasing at #x=-pi/4   #?

Emilia Aguilar · Accepted Answer

Increasing.

The derivative can tell you where an original function is increasing or decreasing. If the derivative is positive at your point of interest, then the original function is increasing at that point. If the derivative is negative at your point of interest, then the original function is decreasing at that point.

Step 1. Determine the derivative of #f(x)#

Requires the product rule

#d/dx f(x)=4x d/dx(cos(3x-(5pi)/4))+cos(3x-(5pi)/4) d/dx(4x)#

#=-12xsin(3x-(5pi)/4)+4cos(3x-(5pi)/4)#

Step 2. Determine if #d/dx f(x)# is positive or negative at the desired point, #x=-pi//4#

At the value #x=-pi/4#, the derivative is positive, which means that the original function, #f(x)# is increasing at that point.

Elijah Abram · Answer

undefined To determine whether $ f(x) = 4x \cos(3x - \frac{5\pi}{4}) $ is increasing or decreasing at $ x = -\frac{\pi}{4} $, we need to evaluate the derivative of the function $ f'(x) $ and then substitute $ x = -\frac{\pi}{4} $ into it.

The derivative of $ f(x) $ with respect to $ x $, denoted as $ f'(x) $, can be found using the product rule and chain rule of differentiation.

$ f'(x) = 4\cos(3x - \frac{5\pi}{4}) - 12x\sin(3x - \frac{5\pi}{4}) $

Now, substitute $ x = -\frac{\pi}{4} $ into $ f'(x) $ to determine whether the function is increasing or decreasing at that point.

$ f'(-\frac{\pi}{4}) = 4\cos(\frac{3\pi}{4} - \frac{5\pi}{4}) - 12(-\frac{\pi}{4})\sin(\frac{3\pi}{4} - \frac{5\pi}{4}) $

$ = 4\cos(-\pi) + 3\pi\sin(\pi) $

$ = 4(-1) + 3\pi(0) $

$ = -4 $

Since $ f'(-\frac{\pi}{4}) = -4 $, which is negative, the function is decreasing at $ x = -\frac{\pi}{4} $.