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

Answer 1

Luke Alvarez

Use the development of the Taylor series of the function

#f(x) = f(x_0)+f'(x_0)(x-x_0) + o((x-x_0^2)#

#sin(2x) ~= sin(pi/3) +2cos(pi/3)(x-pi/6)#

#sin(2x) ~= sqrt(3)/2+(x-pi/6)#

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Answer 2

Cameron Alexander

To find the linearization of ( f(x) = \sin(2x) ) at ( x = \frac{\pi}{6} ), we first find the derivative of ( f(x) ) which is ( f'(x) = 2\cos(2x) ). Then, we evaluate ( f(\frac{\pi}{6}) ) and ( f'(\frac{\pi}{6}) ). ( f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} ) and ( f'(\frac{\pi}{6}) = 2\cos(\frac{\pi}{3}) = -1 ).

So, the equation of the tangent line at ( x = \frac{\pi}{6} ) is ( y = -1(x - \frac{\pi}{6}) + \frac{\sqrt{3}}{2} ). This is the linear approximation of ( f(x) ) 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.