Question

please help! Calculus is going to be the death of me.
find the linear approximation of the function at a. f(x)=sinx a=pi/6

I am suppose to use square root 3/2, 1/2 instead of pi/6

Answer 1

is your question to find the linear approximation for f(1/2) ???

Answer 2

if it is, then you can use the equation of the tangent line at x=pi/6 to approximate f(1/2)=sin(1/2)

Answer 3

well i know the formula is L(x)=f(a)+f'(a)(x-a) do i just plug in 1/2, and why do I use 1/2 instead of square root3/2

Answer 4

f(a)=sin(pi/6) = 1/2
f'(a)=cos(pi/6) = \(\sqrt3/2 \)
a = pi/6
L(x) is the equation of the tangent line at x=pi/6 (in slope-intercept form)

Answer 5

i see...
the tangent line is the linear approximation of f(x)=sinx at x=pi/6......

Answer 6

that makes a lot of sense, I was making it a lot more complicated than it needed to be. Thanks for helping me understand.

Answer 7

yw.... :)