# What is the slope of the line normal to the tangent line of #f(x) = sin(3x-pi) # at # x= pi/3 #?

By the chain rule,

We now determine the slope of the tangent.

The normal line is perpendicular to the tangent, so the slope will be the negative reciprocal.

We can now determine the equation.

Hopefully this helps!

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To find the slope of the line normal to the tangent line of ( f(x) = \sin(3x - \pi) ) at ( x = \frac{\pi}{3} ), we first need to find the slope of the tangent line at ( x = \frac{\pi}{3} ). The slope of the tangent line at a given point is equal to the derivative of the function evaluated at that point.

First, find the derivative of ( f(x) ): [ f'(x) = 3\cos(3x - \pi) ]

Evaluate ( f'(x) ) at ( x = \frac{\pi}{3} ): [ f'\left(\frac{\pi}{3}\right) = 3\cos\left(3\left(\frac{\pi}{3}\right) - \pi\right) = 3\cos(\pi - \pi) = 3\cos(0) = 3 ]

So, the slope of the tangent line at ( x = \frac{\pi}{3} ) is ( m_{\text{tangent}} = 3 ).

The slope of the line normal to the tangent line is the negative reciprocal of the tangent line's slope. Therefore, the slope of the line normal to the tangent line is: [ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{3} ]

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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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