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How do you find the linearization at a=pi/6 of #f(x)=sin2x#?

Answer 1

Samantha Adkison

#L(x) = sqrt3/2+(x-pi/6)#

#L(x) = f(a)+f'(a)(x-a)#

#f(pi/6) = sin(pi/3) = sqrt3/2#

#f'(x) = 2cos2x#, so #f'(pi/6) = 2cos(pi/3) = 1#

#L(x) = sqrt3/2+(x-pi/6)#

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Answer 2

Isabella Ackerman

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 point-slope 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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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.