Home > Calculus > Using the Tangent Line to Approximate Function Values

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

Answer 1

Aurora Alvarado

#L(x) = 1/2+sqrt3/2(x-pi/6)#

#f(x) = sinx#, so #f'(x) = cosx#

At #a = pi/6#, we have #y = f(pi/6) = 1/2# and #f'(pi/6) = sqrt3/2#.

The linearization is the tangent line. So

#L(x) = f(a)+f'(a)(x-a) = 1/2+sqrt3/2(x-pi/6)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isabella Ackerman

To find the linearization of ( f(x) = \sin(x) ) at ( a = \frac{\pi}{6} ), follow these steps:

Compute the first derivative of ( f(x) = \sin(x) ), which is ( f'(x) = \cos(x) ).
Evaluate ( f'(a) ), where ( a = \frac{\pi}{6} ). So, ( f'\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} ).
The linearization of ( f(x) ) at ( a ) is given by the equation ( L(x) = f(a) + f'(a)(x - a) ).
Substitute ( f(a) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} ) and ( f'(a) = \frac{\sqrt{3}}{2} ) into the equation to get ( L(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.