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

Answer 1

Increasing

To find if a function #f(x)# is increasing or deceasing at a point #f(a)#, we take the derivative #f'(x)# and find #f'(a)#/

If #f'(a)>0# it is increasing If #f'(a)=0# it is an inflection If #f'(a)<0# it is decreasing

#f(x)=cosx+sinx#

#f'(x)=-sinx+cosx#

#f'(pi/6)=cos(pi/6)-sin(pi/6)=(-1+sqrt(3))/2#

#f'(pi/6)>0#, so it is increasing at #f(pi/6)#

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Answer 2

Scarlett Allison

To determine if ( f(x) = \cos(x) + \sin(x) ) is increasing or decreasing at ( x = \frac{\pi}{6} ), we need to analyze its derivative.

( f'(x) = -\sin(x) + \cos(x) )

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

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

( = -\frac{1}{2} + \frac{\sqrt{3}}{2} )

( = \frac{\sqrt{3}}{2} - \frac{1}{2} )

Since ( \frac{\sqrt{3}}{2} - \frac{1}{2} > 0 ), ( f'(x) ) is positive at ( x = \frac{\pi}{6} ). Therefore, ( f(x) ) is increasing at ( x = \frac{\pi}{6} ).

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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.