Is #f(x)=cot(2x)*tanx# increasing or decreasing at #x=pi/6#?

Answer 1

#f(x)# is decreasing at #x=pi/6#

A function's first derivative at a given point determines whether the function is increasing or decreasing at that point.

As #f(x)=cot2x*tanx=1/(tan2x) xxtanx#
#=(1-tan^2x)/(2tanx)xx2tanx=(1-tan^2x)/2=1/2-1/2tan^2x#
Hence, #f'(x)=(df)/(dx)=-1/2 xx 2tanx xx sec^2x=-tanxsec^2x#
and #f'(pi/6)=-tan(pi/6)sec^2(pi/6)=-1/sqrt3 xx (2/sqrt3)^2=-4/(3sqrt3)#
As #f'(pi/6) < 0#, #f(x)# is decreasing at #x=pi/6#
Observe that #x=pi/6=0.5236# the function is decreasing in the graph below.

graph{tanx [-2.5, 2.5, -1.25, 1.25]} \cot(2x)

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Answer 2

To determine if ( f(x) = \cot(2x) \cdot \tan(x) ) is increasing or decreasing at ( x = \frac{\pi}{6} ), we need to evaluate the derivative of ( f(x) ) at that point.

The derivative of ( f(x) ) can be found using the product rule and chain rule:

[ f'(x) = \frac{d}{dx} [\cot(2x)] \cdot \tan(x) + \cot(2x) \cdot \frac{d}{dx}[\tan(x)] ]

[ f'(x) = -2\csc^2(2x) \cdot \tan(x) + \cot(2x) \cdot \sec^2(x) ]

Now, plug in ( x = \frac{\pi}{6} ) to find the value of the derivative:

[ f'\left(\frac{\pi}{6}\right) = -2\csc^2\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) + \cot\left(\frac{\pi}{3}\right) \cdot \sec^2\left(\frac{\pi}{6}\right) ]

[ f'\left(\frac{\pi}{6}\right) = -2\csc^2\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) + \cot\left(\frac{\pi}{3}\right) \cdot \sec^2\left(\frac{\pi}{6}\right) ]

[ f'\left(\frac{\pi}{6}\right) = -2\csc^2\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) + \cot\left(\frac{\pi}{3}\right) \cdot \sec^2\left(\frac{\pi}{6}\right) ]

[ f'\left(\frac{\pi}{6}\right) = -2\left(\frac{2}{\sqrt{3}}\right)^2 \cdot \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2} \cdot \left(\frac{2}{\sqrt{3}}\right)^2 ]

[ f'\left(\frac{\pi}{6}\right) = -\frac{8}{3} + \frac{\sqrt{3}}{2} \cdot \frac{2}{3} ]

[ f'\left(\frac{\pi}{6}\right) = -\frac{8}{3} + \frac{2\sqrt{3}}{3} ]

[ f'\left(\frac{\pi}{6}\right) = -\frac{8}{3} + \frac{2\sqrt{3}}{3} ]

[ f'\left(\frac{\pi}{6}\right) = -\frac{8-2\sqrt{3}}{3} ]

Since ( f'\left(\frac{\pi}{6}\right) < 0 ), ( f(x) ) is decreasing 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.

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.

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.

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.

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