Is #f(x)= x-12sinx # increasing or decreasing at #x=-pi/6 #?

Answer 1

The function is decreasing.

If #f'(-pi/6)<0#, then the function is decreasing at #x=-pi/6#. If #f'(-pi/6)>0#, then the function is increasing at #x=-pi/6#.

Find #f'(x)#.

#f'(x)=1-12cosx#

Find #f'(-pi/6)#.

#f'(-pi/6)=1-12cos(-pi/6)=1-12(sqrt3/2)=1-6sqrt3#

Since #f'(-pi/6)~~-9.39<0#, the function is decreasing at #x=-pi/6#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Arocha

To determine whether the function ( f(x) = x - 12 \sin(x) ) is increasing or decreasing at ( x = -\frac{\pi}{6} ), we need to examine the sign of its derivative at that point.

First, let's find the derivative of ( f(x) ) using the sum and chain rules:

[ f'(x) = 1 - 12 \cos(x) ]

Now, evaluate ( f'(-\frac{\pi}{6}) ):

[ f'(-\frac{\pi}{6}) = 1 - 12 \cos(-\frac{\pi}{6}) = 1 - 12 \cdot \frac{\sqrt{3}}{2} = 1 - 6\sqrt{3} ]

Since ( 1 - 6\sqrt{3} ) is negative, the function ( f(x) = x - 12 \sin(x) ) is decreasing at ( x = -\frac{\pi}{6} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.