How do you find the linearization at a=pi/6 of #f(x)=sin2x#?
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To find the linearization of ( f(x) = \sin(2x) ) at ( a = \frac{\pi}{6} ), follow these steps:

Find the first derivative of ( f(x) ), denoted ( f'(x) ). [ f'(x) = 2\cos(2x) ]

Evaluate ( f'(a) ) at ( a = \frac{\pi}{6} ). [ f'\left(\frac{\pi}{6}\right) = 2\cos\left(2\left(\frac{\pi}{6}\right)\right) = 2\cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 ]

Write down the linearization formula using the pointslope form: [ L(x) = f(a) + f'(a)(x  a) ]

Substitute the values of ( f(a) ), ( f'(a) ), and ( a ) into the formula. [ L(x) = \sin(2\cdot\frac{\pi}{6}) + 1\cdot(x  \frac{\pi}{6}) ]

Simplify the expression. [ L(x) = \sin\left(\frac{\pi}{3}\right) + (x  \frac{\pi}{6}) ]

Finalize the answer. [ L(x) = \frac{\sqrt{3}}{2} + x  \frac{\pi}{6} ]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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